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ivanzaharov [21]
2 years ago
7

ABCD is a trapezoid. Find y

Mathematics
2 answers:
aleksandrvk [35]2 years ago
5 0
Y = 105 degrees

Angle A = Angle B
Angle C = Angle D
Trapezoid: 360 degrees

75+75= 150
360 - 150 = 210
210/2 = 105
babunello [35]2 years ago
4 0

Answer:

C

Step-by-step explanation:

\angle A + \angle D = 180

\angle A + y = 180

180 = 75 + y

y = 180 - 75 = 105

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kvasek [131]
So each house can hold 1000 cubes that are 1 meter in length. The house is also shaped like a cube, so you need to cube-root 1000. The cube-root of 1000 is 10. So the cube house has a length, width, and height of 10 meters.
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3 years ago
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If f x is a linear function, f(-1)=3, and f(1)=5, find an equation for f(x)
BartSMP [9]

The equation of f(x) is f(x) = x + 4

<h3>How to determine the equation?</h3>

The given parameters are

f(-1) = 3 and f(1) = 5

Start by calculating the slope (m) using:

m = \frac{f(1) - f(-1)}{1 - -1}

This gives

m = \frac{5 - 3}{1 + 1}

Evaluate

m = 1

The equation is then calculated as;

f(x) = m(x - x1) + f(-1)

This gives

f(x) = 1(x + 1) + 3

Expand

f(x) = x + 4

Hence, the equation of f(x) is f(x) = x + 4

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brainly.com/question/4025726

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1 year ago
The cake store is having a 25% off sale on all of its cakes. If the cake you want regularly costs $9 how much would you save wit
Gre4nikov [31]
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One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

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c^2 = 9dp

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8 0
3 years ago
Consider the volume of a rectangular pyramid, one whose base is not square. Does changing one of the three variables (length, wi
user100 [1]

Let

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we know that

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so

the three variables are going to have the same influence on the volume


example

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the new volume is equal to

V new=(1/3)*[(L/2)*W]*H-----> V new=(1/2)*V

the new volume is half the original volume


case b) let's change the variable W (we reduce the variable by half)

W=(1/2)*W

the new volume is equal to

V new=(1/3)*[(W/2)*L]*H-----> V new=(1/2)*V

the new volume is half the original volume


case c) let's change the variable H (we reduce the variable by half)

H=(1/2)*H

the new volume is equal to

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the new volume is half the original volume


therefore


the answer is

the three variables have the same influence on the volume

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