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3 years ago

####I NEED HELP PLEASE#### Will give brainlist!!!!!!

Mathematics

1 answer:

hoa [83]3 years ago

4 0

Answer:

9.03

Step-by-step explanation:

Just divide 63.21 by 7

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What is the postulate theorem for these triangles?

Inessa [10]

Answer:

1) AAS

2) SSS

Step-by-step explanation:

1) ANGLE ANGLE SIDE POSTULATE

2) SIDE SIDE SIDE POSTULATE

6 0

3 years ago

Ava is saving for A new computer that cost 1218 she has already saved half of the money Ava earns $14 per hour how many hours mu

11Alexandr11 [23.1K]

609 + 14x = 1218
14x = 609
x = 43.5 hours

4 0

4 years ago

Read 2 more answers

What is the vertex form of 2x^2+12x+14

Finger [1]

Answer:

2 (x + 3)^2 - 2

Step-by-step explanation:

1. Factor the 2 out of the equation (don't forget about it - it will be used later). This gives you x^2 + 6x + 7

2. Subtract the 7 over. This gives you x^2 + 6x = -7

3. Take b (6), divide it by 2 (3), and square it (9). Then add that to both sides of the equation. This gives you x^2 + 6x + 9 = 2

4. Factor the left side of the equation. This gives you (x + 3)^2 = 2

5. Subtract the 2 over. This gives you (x + 3)^2 - 2

6. Remember that 2 we factored out of the equation earlier? Well that goes in front of the parentheses now, giving you the final answer of 2 (x + 3)^2 -2

8 0

4 years ago

A quadrilateral with vertices C(−6, 6), D(−1, 10), E(5, 8), F(0, 4) is a parallelogram.

Usimov [2.4K]

Answer:

m = (y2 - y1) / (x2 - x1) = (10 - 6) / (-1 - (-6)) = 4 / 5 = 0.8

4 0

3 years ago

Find the measures of the three angles, in radians, of the triangle with the given vertices: d(1,1,1), e(1,−5,2), and f(−2,2,7).

Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

$\overrightarrow{DE}=(1-1,-5-1,2-1)=(0,-6,1),\\ \\\overrightarrow{DF}=(-2-1,2-1,7-1)=(-3,1,6).$

Then

$\cos \angle D=\dfrac{0\cdot (-3)+(-6)\cdot 1+1\cdot 6}{\sqrt{0^2+(-6)^2+1^2}\cdot \sqrt{(-3)^2+1^2+6^2}}=\dfrac{0}{\sqrt{37} \cdot \sqrt{46} }=0.$

2. Find

$\overrightarrow{ED}=(1-1,1-(-5),1-2)=(0,6,-1),\\ \\\overrightarrow{EF}=(-2-1,2-(-5),7-2)=(-3,7,5).$

Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

3. Find

$\overrightarrow{FE}=(1-(-2),-5-2,2-7)=(3,-7,-5),\\ \\\overrightarrow{FD}=(1-(-2),1-2,1-7)=(3,-1,-6).$

Then

$\cos \angle F=\dfrac{3\cdot 3+(-7)\cdot (-1)+(-5)\cdot (-6)}{\sqrt{3^2+(-7)^2+(-5)^2}\cdot \sqrt{3^2+(-1)^2+(-6)^2}}=\dfrac{46}{\sqrt{83} \cdot \sqrt{46} }=\sqrt{\dfrac{46}{83}}.$

$\angle E=\arccos0=\dfrac{\pi}{2},\\ \\
\angle D=\arccos\letf(\sqrt{\dfrac{37}{83}}\right)\approx 0.27\pi,\\ \\
\angle F=\arccos\letf(\sqrt{\dfrac{46}{83}}\right)\approx 0.23\pi.$

3 0

3 years ago