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2 years ago

Find the surface area of the rectangular prism

Mathematics

2 answers:

Nezavi [6.7K]2 years ago

7 0

The answer is 158 the length times width times height

Lapatulllka [165]2 years ago

5 0

Answer:

158 in

Step-by-step explanation:

front/back faces: 8 * 3 = 24in

side faces: 8 * 5 = 40in

top/bottom faces: 3 * 5 = 15in

2 * (24 + 40 + 15) = 158in

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The equation $(x - 7)^2 + (y + 13)^2 = 12$ represents a circle. What is its radius?

miskamm [114]

Answer:

Step-by-step explanation:

The standard equation of a circle is expressed as;

(x-a)²+(y-b)² = r² where;

(a, b) is the center of the circle

r is the radius of the circle

Given the equation (x - 7)^2 + (y + 13)^2 = 12

On comparing;

r² = 12

Take the square root of both sides

√r² = √12

r = √12

r = √4*3

r = 2√3

Hence the radius of the circle is 2√3

7 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

3 years ago

Climbing at an indoor rock-climbing gym costs $8 per hour. Lizbeth has a $30 gift card to spend at the gym. Write an algebraic e

Aleks04 [339]

Answer:

8h=30

Lizbeth can not climb for 4.5 hours.

Step-by-step explanation:

If you plug in 4.5 for h you get

8(4.5)=30

36 does not equal 30

Therefore Lizbeth can't climb for 4.5 hours

6 0

3 years ago

For f(x)= 4x+1 and g(x) = x squared -5 , find (f-g)(x)

makvit [3.9K]

Hope i am not wrong
f(x)=4x+1
g(x)=x^2-5

(f-g)(x)=f(x)-g(x)
(f-g)(x) =4x+1-x^2+5

-(x^2-6-4x)
solve it for x

6 0

3 years ago

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The distance you drive is proportional to the amount of gas used. Driving 132 miles uses 4 gallons of gas. Complete parts (a) t

Goshia [24]

Answer:

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3 years ago