A______ is a value, usually in the form of a letter or symbol, that can change or that represents an unknown quantity.

Answer: 1

Step-by-step explanation: 863x14ysqrt9= 3 simplified is 1

Answer:

B. 61

Step-by-step explanation:

Given:

∆PQR ≅ ∆PQS

PQ = 2x + 41

QS = 7x - 24

QR = 3x + 16

Required:

Numerical value of PQ

SOLUTION:

First, create an equation to find the value of x as follows:

Since both triangles are congruent, therefore:

QS = QR

7x - 24 = 3x + 16 (Substitution)

Collect like terms

7x - 3x = 24 + 16

4x = 40

Divide both sides by 4

4x/4 = 40/4

x = 10

Find PQ by plugging x = 10 into PQ = 2x + 41

PQ = 2(10) + 41

PQ = 20 + 41

PQ = 61

Answer:

y=2

y=0

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    7*y-1-(3*y^2+y-1)=0

(7y - 1) -  ((3y2 +  y) -  1)  = 0  3.1     Pull out like factors :

  6y - 3y2  =   -3y • (y - 2)

Answer:

y = -21/2

Step-by-step explanation:

multiply both sides of the equation by 21 so it would be

7y+42=3y

move the terms

7y-3y=-42

collect like terms

4y=-42

divide both sides by four

y= -21/2

Answer:

the coordinates of b is (9,7)  

the coordinates of a is (-4,-6)

Step-by-step explanation:

Answer: X=6

you would do 30 - 12 witch equals 18

than you would do 18 divided by 3 witch is 6 so it is x = 6

Answer:

Step-by-step explanation:

5,2

that is a rational number.

Answer: The percent markup for a shell necklace = 200%

Step-by-step explanation:

Given: Cost price per necklace = $3.75

Selling price per necklace =  $11.25

Mark up = (Selling price -  Cost price )

= $ (11.25 -3.75 )

=$ 7.5

The percent markup for a shell necklace = [tex]\dfrac{\text{Mark up}}{\text{cost price}}\times100\%[/tex]

[tex]=\dfrac{7.5}{3.75}\times100\%\\\\=2\times100\%=200\%[/tex]

Hence, The percent markup for a shell necklace = 200%

The percent markup for a shell necklace is 200%.

Markup can be defined as the rate at which the cost of a product is increased. The higher the markup, the higher the profit earned by a seller all things being equal.

Percentage markup = (profit / cost price) x 100

Profit = $11.25 - $3.75 = $7.50

Percentage markup = ($7.50 / $3.75) x 100 = 200%

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Answer:

a. 3

Step-by-step explanation:

Answer:

(A)  X = 3

Step-by-step explanation:

-_-

Answer:

y=0x-1

Step-by-step explanation:

Slope-int form: y=mx+b

Both points have the same y value and different x values, so the slope (m) is 0.

Since the y-intercept is (0, -1), b=-1

Answer:

The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Step-by-step explanation:

The net change in 7 students' scores on the exam after completing the course are:

S = {37 ,12 ,12 ,17 ,13 ,32 ,23}

Compute the sample mean and sample standard deviation as follows:

[tex]\bar x=\frac{1}{n}\sum x=\frac{1}{7}\times 146=20.857\\\\s=\sqrt{\frac{1}{n-1}\sum (x-\bar x)^{2}}}=\sqrt{\frac{1}{7}\times 622.8571}=10.189[/tex]

As the population standard deviation is not known, a t-interval will be formed.

Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:

[tex]t_{\alpha/2, (n-1)}=t_{0.20/2, (7-1)}=t_{0.10,6}=1.415[/tex]

*Use a t-table.

Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:

[tex]CI=\bar x\pm t_{\alpha/2, (n-1)}\times\frac{s}{\sqrt{n}}[/tex]

     [tex]=20.857\pm 1.415\times\frac{10.189}{\sqrt{7}}\\\\ =20.857\pm 5.4493\\\\=(15.4077, 26.3063)\\\\\approx (15.4,26.3)[/tex]

Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Using the given definition, for [tex]f(x)=\frac1{\sqrt x}+x[/tex], we have

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\left(\frac1{\sqrt{x+h}}+x+h\right)-\left(\frac1{\sqrt x}+x\right)}h[/tex]

Right away, we see x and -x in the numerator, so we can drop those terms.

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}+h-\frac1{\sqrt x}}h[/tex]

Remember that limits distribute over sums, i.e.

[tex]\displaystyle\lim_{x\to c}(f(x)+g(x))=\lim_{x\to c}f(x)+\lim_{x\to c}g(x)[/tex]

so we can separate the h from everything else in the numerator:

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+\lim_{h\to0}\frac hh[/tex]

Since h ≠ 0, we have [tex]\frac hh=1[/tex], so the second limit is simply 1.

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{\frac1{\sqrt{x+h}}-\frac1{\sqrt x}}h+1[/tex]

For the remaining limit, focus on the numerator for now. Combine the fractions in the numerator:

[tex]\dfrac1{\sqrt{x+h}}-\dfrac1{\sqrt x}=\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}[/tex]

Recall the difference of squares identity,

[tex]a^2-b^2=(a-b)(a+b)[/tex]

Let [tex]a=\sqrt x[/tex] and [tex]b=\sqrt{x+h}[/tex]. Multiply the numerator and denominator by [tex](a+b)[/tex], so that the numerator can be condensed using the identity above.

[tex]\dfrac{\sqrt x-\sqrt{x+h}}{\sqrt x\sqrt{x+h}}\cdot\dfrac{\sqrt x+\sqrt{x+h}}{\sqrt x+\sqrt{x+h}}[/tex]

[tex]=\dfrac{(\sqrt x)^2-(\sqrt{x+h})^2}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

[tex]=\dfrac{x-(x+h)}{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

[tex]=-\dfrac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}[/tex]

Back to the limit: all this rewriting tells us that

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{-\frac h{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}}h+1[/tex]

Again, the h's cancel, and we can pull out the factor of -1 from the numerator and simplify the fraction:

[tex]f'(x)=\displaystyle-\lim_{h\to0}\frac1{\sqrt x\sqrt{x+h}(\sqrt x+\sqrt{x+h})}+1[/tex]

The remaining expression is continuous at h = 0, so we can evaluate the limit by substituting directly:

[tex]f'(x)=-\dfrac1{\sqrt x\sqrt{x+0}(\sqrt x+\sqrt{x+0})}+1[/tex]

[tex]f'(x)=-\dfrac1{2x\sqrt x}+1[/tex]

or, if we write [tex]\sqrt x=x^{1/2}[/tex], we get

[tex]f'(x)=-\dfrac12x^{-3/2}+1[/tex]

Answer:

Yess

Step-by-step explanation:

If you multiply all the numbers together you get 39690000. And the square root of that number is 6300. So it is a perfect square. If you did 6300² = 39690000.

Answer:

7 <  3N+1 < 11

6 < 3n < 10

2 < n < 10/3

Step-by-step explanation:

The required number is -3, -2, -1, 0, and 1 which is three times a number, increased by one, is between negative eleven and seven.

Given that,
Three times a number, increased by one, is between negative eleven and seven. the number is to be determined.

Inequality can be defined as the relation of the equation containing the symbol of ( ≤, ≥, <, >) instead of the equal sign in an equation.

here,
Accoding to the condition let the number be x,
and
-11 < 3x + 1 < 7
-12 < 3x < 6
-4 < x < 2
So the number are -3, -2, -1, 0, anb 1

Thus, the required number is -3, -2, -1, 0, and 1 which is three times a number, increased by one, is between negative eleven and seven.

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Answer:

Option (A)

Step-by-step explanation:

Given expression is,

[tex]5\sqrt{48a}=60[/tex]

Squaring on both the sides of the equation,

[tex](5\sqrt{48a})^2=(60)^2[/tex]

25(48a) = 3600

1200a = 3600

By dividing the equation by 1200,

a = [tex]\frac{3600}{1200}[/tex]

a = 3

Therefore, Option (A) will be the answer.

Answer:

40?

Step-by-step explanation:

it literally tells you?

Answer:

90 because 65 plus 25 is 90.

Step-by-step explanation:

65 plus 25 is 90 and

Answer:

62.5

Step-by-step explanation:

There ya gooo :)

Answer:

62.5%

Here You go

Step-by-step explanation:

Answer:

Step-by-step explanation:

Area of rectangle= Length x Width

and the given Area is a quadratic expression

A=[tex]x^{2} -4x-21[/tex]

We use the factorization method so,

we need 2 numbers that when multiplied we get -21 and when we add/subtract we get -4 so,

A=[tex]x^{2} -7x+3x-21[/tex]

now we simplify,

A=[tex]x(x-7)+3(x-7)[/tex]

A=[tex](x-7)(x+3)[/tex]

This looks familiar doesn't it, when we write the formula for the area of rectangle its

A= Length x Width and the equation here shows that

A= [tex](x-7)(x+3)[/tex]

So the expression for the length is x-7 and

the expression for the width is x+3 i think u have missed maybe some information on the question as such that the perimeter might be missing because length could be either x-7 or even x+3 same goes for width maybe someone can correct me if im wrong

Answer:

-2x+3

Step-by-step explanation:

Using y=mx+b

To make the line perpendicular, you just need to make the slope (m=2) negative. To make it go through (-1, 5), you just need to adjust the b value to raise the graph to a point where it will pass through the point.

Answer:  Think about diagonal sides and horizontal bases.

Step-by-step explanation:  "Think outside of the box." as they say.

See the screenshot attached.

Answer:

i need a picture

Step-by-step explanation:

Answer:

Option A

Step-by-step explanation:

The answer is option A "A number less than 2." That is because we check off each of the following given options. It won't be option C because they're 4 odd numbers a equal amount of multiplies of two (option D) which means they both have a equal (4 out of 8) chance. It won't be option B because they're 7 out of 8 numbers that are greater then on which means you have a greater chance of getting a number greater then one but the question is asking for the most unlikely.....therefore the answer is option A because you have a 1 out of 8 chance of getting a number less then two.

Hope this helps.

Answer:

21 quarters and 19 dimes

Step-by-step explanation:

Create a system of equations where q is the number of quarters and d is the number of dimes.

0.25q + 0.1d = 7.15

q + d = 40

Solve by elimination by multiplying the bottom equation by -0.25

0.25q + 0.1d = 7.15

-0.25q - 0.25d = -10

Add them together and solve for d:

-0.15d = -2.85

d = 19

Then, plug in 19 as d into the second equation to solve for q:

q + d = 40

q + 19 = 40

q = 21

So, there are 21 quarters and 19 dimes

Answer:

21 Quarters, 19 Dimes.

Step-by-step explanation:

21 x .25 = 5.25

19 x .10 = 1.90

5.25 + 1.90 = 7.15

I guess you have to find

[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}[/tex]

given that [tex]\tan\theta=\frac8{15}[/tex].

We can immediately solve for [tex]\sec\theta[/tex]:

[tex]\sec^2\theta=1+\tan^2\theta\implies\sec\theta=\pm\dfrac{17}{15}[/tex]

(without knowing anything else about [tex]\theta[/tex], we cannot determine the sign)

Then we get [tex]\cos\theta[/tex] for free:

[tex]\cos\theta=\dfrac1{\sec\theta}=\pm\dfrac{15}{17}[/tex]

and we can now solve for [tex]\sin\theta[/tex]:

[tex]\sin^2\theta+\cos^2\theta=1\implies \sin\theta=\pm\dfrac8{17}[/tex]

Notice that we have 2*2 = 4 possible choices of sign for either sin or cos.

• If both are positive, then

[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{391}{90}[/tex]

• If both are negative, then

[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{391}{480}[/tex]

• If sin is positive and cos is negative, then

[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{119}{480}[/tex]

• If cos is positive and sin is negative, then

[tex]\dfrac{\sin\theta+\cos\theta}{\cos\theta(1-\cos\theta)}=\dfrac{119}{30}[/tex]

Answer:

Step-by-step explanation:

answer:

48p² + 4p - 4

solution:

8p × 6p +8p × 2 - 2 × 6p - 2 × 2

48p² + 8p x 2 - 2 ×6p - 2 × 2

48p² + 16p - 12p - 4

48p² + 4p - 4

The required algebraic product is 48[tex]p^{2}[/tex] + 4p - 4.

Given the two binomial (8p - 2)(6p + 2).

To multiply two binomials, take the first term of the first binomial and multiply with the entire second binomial and positive or negative as per given in the first binomial ,take the second term of the first binomial and multiply with the entire second binomial.

Let a, b, c and d be any four variables. Consider (a - b)(c + d) gives

a(c + d) - (c + d).

That implies, (8p - 2)(6p + 2) = 8p(6p + 2) - 2(6p + 2)

Multiply by removing the brackets gives,

(8p - 2)(6p + 2) = 48[tex]p^{2}[/tex] + 16p - 12p - 4

Combining like terms and algebraic sum gives,

(8p - 2)(6p + 2) = 48[tex]p^{2}[/tex] + 4p - 4.

Hence, the required algebraic product is 48[tex]p^{2}[/tex] + 4p - 4.

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