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3 years ago

PLEASE HELP EASY 6TH GRADE MATH! WHOEVER DOES THE BEST GETS BRAINLIEST PLZZZZZZ?

Mathematics

1 answer:

zhuklara [117]3 years ago

8 0

Answer:

1.B 2.B 3.A

Step-by-step explanation:

1.(70-24)/2 = 23

2. 295 x 8.95 =2640.25 - 400 = 2240.25 ~ $2000

3. 21hoursx48 weeks x 2years = 2016 hours ~ 2000

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Please answer my question thank you.

Andrews [41]

A. terminating
b. repeating
c. repeating
d. terminating
e. repeating
i think

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3 years ago

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5x-15y=45 solve for y

shepuryov [24]

Answer:

y=(0,-3)

Step-by-step explanation:

To find the y-intercept, substitute in 0 for x and solve for y

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2 years ago

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Z4 = Z5 so A//B<br> True or <br> False

Jet001 [13]

Answer:

True

Step-by-step explanation:

∠4 ≅ ∠5 are corresponding angle and equal to each other

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2 years ago

What is the answer to this??

Firdavs [7]

Answer:

x=3

z=65

Step-by-step explanation:

6x+97 and 14x+73 are vertical angles which means they are equal

6x+97= 14x+73

Subtract 6x from each side

6x-6x+97= 14x-6x+73

97 = 8x +73

Subtract 73 from each side

97-73 = 8x+73-73

24 = 8x

Divide each side by 8

24/8 = 8x/8

3 = x

Now we need to find z

6x+97 and z are supplementary angles which means they add to 180

6x+97 +z = 180

6(3) +97 +z = 180

18+97+z=180

Combine like terms

115+z = 180

Subtract 115 from each side

115-115 +z=180-115

z = 65

3 0

3 years ago

100 POINTS AND BRAINLIEST

zysi [14]

Answer:

8 square units and $\frac{40}{3}$ square units

Step-by-step explanation:

The area of the triangle ABC is 24 square units.

1. Triangles ABC and FBG are similar with scale factor $\frac{1}{3},$ then

$\dfrac{A_{\triangle FBG}}{A_{\triangle ABC}}=\dfrac{1}{9}\Rightarrow A_{\triangle FBG}=\dfrac{1}{9}\cdot 24=\dfrac{8}{3}\ un^2.$

2. Triangles ABC and DBE are similar with scale factor $\frac{2}{3},$ then

$\dfrac{A_{\triangle DBE}}{A_{\triangle ABC}}=\dfrac{4}{9}\Rightarrow A_{\triangle DBE}=\dfrac{4}{9}\cdot 24=\dfrac{32}{3}\ un^2.$

3. Thus, the area of the quadrilateral DFGE is

$A_{DFGE}=A_{\triangle DBE}-A_{\triangle FBG}=\dfrac{32}{3}-\dfrac{8}{3}=8\ un^2.$

and the area of the quadrilateral ADEC is

$A_{ADEC}=A_{\triangle ABC}-A_{\triangle DBE}=24-\dfrac{32}{3}=\dfrac{40}{3}\ un^2.$

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3 years ago

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