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Zolol [24]
3 years ago
13

Terri wrote this equation for the area of a parallelogram.

Mathematics
1 answer:
Helga [31]3 years ago
7 0
220.44=16.7h
220.44/16.7=h
h=13.2
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Jace gathered the data in the table. He found the approximate line of best fit to be y = –0.7x + 2.36.
MaRussiya [10]
The residual value is -1.14.
Plug 5 into x
y=-0.7(5)+2.36
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3 0
3 years ago
Can you help me please?! :)
UkoKoshka [18]
The answer is A
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the rest of the work is shown here

3 0
2 years ago
HELP! Simplify: -6 (5/3x+4)<br> A) -10x-24. B) -10x-24. C) -10x+4. D) -10x-4.
pashok25 [27]

Answer:

D) -10x-4

Step-by-step explanation:

−6(5/3x+4)

=(−6)(5/3x+4)

=(−6)(5/3x)+(−6)(4)

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6 0
3 years ago
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I will mark brainliest if you are correct!
notsponge [240]

Answer:

i think the answer is C

Step-by-step explanation:

6 0
3 years ago
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The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.
ehidna [41]
1. Check the drawing of the rhombus ABCD in the picture attached.

2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.

3. The diagonals:
     i) bisect the angles so m(ODC)=60°/2=30°

     ii) are perpendicular to each other, so m(DOC)=90°

4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.

5. By the pythagorean theorem, DO= \sqrt{ DC^{2}- OC^{2} }=  \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}= 
=\sqrt{9}* \sqrt{3} =3\sqrt{3}  (in)

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}==2*9 \sqrt{3}=18 \sqrt{3} (in^{2})

6 0
3 years ago
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