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3 years ago

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Which of the following is a step in simplifying the expression x multiplied by y to the power of 3 over x to the power of negati

ve 4 multiplied by y to the power of 4, the whole to the power of negative 2.?

1 answer:

stich3 [128]3 years ago

5 0

Answer:

-3

Step-by-step explanation:

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In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an

Airida [17]

This is a little long, but it gets you there.

ΔEBH ≅ ΔEBC . . . . HA theorem
EH ≅ EC . . . . . . . . . CPCTC
∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
ΔDAC ≅ ΔDAG . . . HA theorem
DC ≅ DG . . . . . . . . . CPCTC
∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
(∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)

5 0

3 years ago

1. 15<br> 2. 27<br> 3. 75<br> 4. 87

masya89 [10]

I’m 15 so the answer is number 1

3 0

3 years ago

Let R be the triangular region in the first quadrant, with vertices at points (0,0), (0,2), and (1,2). The region R is the base

s2008m [1.1K]

The line through (0, 0) and (1, 2) has equation $y=2x$ .

Each cross section has a leg in the $xy$ -plane whose length is the horizontal distance from the $y$ -axis to the line $y=2x$ , which is $x=\frac y2$ .

The cross sections are isosceles right triangles, so the legs that lie perpendicular to the $xy$ -plane have the same length as the legs *in* the $xy$ -plane, hence these triangles haves bases and heights equal to $\frac y2$ . Then the area of each cross section is

$\dfrac12\left(\dfrac y2\right)^2=\dfrac{y^2}8$

where the cross sections are generated over the interval $0\le y\le2$ .

The volume of the solid is then

$\displaystyle\int_0^2\frac{y^2}8\,\mathrm dy=\frac1{24}y^3\bigg|_0^2=\frac{2^3}{24}=\frac13$

8 0

3 years ago

How you express the average of five quizzes is 85 in mathematical expression

zvonat [6]

Let x1, x2, x3, x4, x5 be the scores for the five tests respectively and y be the average score of those tests.

y = (x1 + x2 + x3 + x4 + x5) / 5

since y = 85:

(x1 + x2 + x3 + x4 + x5) / 5 = 85

4 0

3 years ago

Find the perimeter and the area of the regular polygon

Andreas93 [3]

The perimeter of the hexagon is 180 ft.

The area of the hexagon is 2338.3 ft

<h3>How to find area and perimeter of a polygon?</h3>

The perimeter of the polygon can be found as follows:

The polygon is a regular hexagon.

Therefore,

perimeter of the hexagon = 30 × 6 = 180 ft

Area of the hexagon = 3√3 / 2 × s²

where

s = side length

Therefore,

Area of the hexagon = 3√3 / 2 × 30²

Area of the hexagon = 3√3 / 2 × 900

Area of the hexagon = 450 × 3√3

Area of the hexagon = 2338.26859022

Area of the hexagon = 2338.3 ft

learn more on polygon here: brainly.com/question/4014163

#SPJ1

5 0

2 years ago