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3 years ago

3. Solve this inequality: 3b-7<32 O A.b<39 B.b<5 C.b< 13 D.b<8.33

Mathematics

1 answer:

pentagon [3]3 years ago

7 0

Answer:

C.b< 13

Step-by-step explanation:

3b-7<32

Add 7 to each side

3b-7+7<32+7

3b < 39

Divide each side by 3

3b< 39/3

b < 13

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Answer:

Step-by-step explanation:

20) Angles 1 and 2 are supplementary angles( sum of angle 1 and angle 2 is 180 degrees)

21) Angles 1 and 2 are supplementary angles( sum of angle 1 and angle 2 is 180 degrees)

22) Angle TUW + Angle WUV =Angle TUV

Therefore

(7x-9) + (5x - 11) = (9x+1)

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23) GEF - 13 = 5DEG

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2 years ago

RectangleABCD has vertices at A(– 3, 1),B(– 2, – 1),C(2, 1), andD(1, 3). What is the area, in square units, of this rectangle? A

Anna [14]

Answer:

Option A. $10\ units^{2}$

Step-by-step explanation:

we know that

The area of the rectangle is equal to

A=LW

where

L is the length of rectangle

W is the width of rectangle

we have

$A(-3,1),B(-2,-1),C(2,1),D(1,3)$

Plot the vertices

see the attached figure

L=AD=BC

W=AB=DC

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

<em>Find the distance AD</em>

$A(-3,1),D(1,3)$

substitute in the formula

$AD=\sqrt{(3-1)^{2}+(1+3)^{2}}$

$AD=\sqrt{(2)^{2}+(4)^{2}}$

$AD=\sqrt{20}$

$AD=2\sqrt{5}\ units$

<em>Find the distance AB</em>

$A(-3,1),B(-2,-1)$

substitute in the formula

$AB=\sqrt{(-1-1)^{2}+(-2+3)^{2}}$

$AB=\sqrt{(-2)^{2}+(1)^{2}}$

$AB=\sqrt{5}$

$AB=\sqrt{5}\ units$

Find the area

$A=(2\sqrt{5})*(\sqrt{5})=10\ units^{2}$

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