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2 years ago

Lines of symmetry give e the answer

Mathematics

1 answer:

quester [9]2 years ago

6 0

Answer:

Step-by-step explanation:

There are 4 reflectional symmetry

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Solve the system of equations -x+ 2y = –11 and 5x – 8y = 39 by combining the equations.

Alchen [17]

Answer:

x=-5, y=-8. (-5, -8).

Step-by-step explanation:

-x+2y=-11

5x-8y=39

---------------

5(-x+2y)=5(-11)

5x-8y=39

---------------------

-5x+10y=-55

5x-8y=39

--------------------

2y=-16

y=-16/2

y=-8

-x+2(-8)=-11

-x-16=-11

-x=-11+16

-x=5

x=-5

3 0

2 years ago

Round to the nearest hundredth? A 4 ? 35° с B

DaniilM [7]

The length of the unknown side length, |AB|, is 6.97

<h3>Trigonometry</h3>

From the question, we are to determine the value of the unknown side length

The unknown side length is the hypotenuse

Using SOH CAH TOA

We can write that

sin 35° = 4/|AB|

∴ |AB| = 4 /sin35°

|AB| = 4 /sin35°

|AB| = 6.97

Hence, the length of the unknown side length, |AB|, is 6.97

Learn more on Trigonometry here: brainly.com/question/22099203

#SPJ1

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1 year ago

F(x) = –3x2 – 7x Find f(7)

Dominik [7]

Answer:

Step-by-step explanation:

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2 years ago

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What is the volume of the triangular prism in cubic centimeters ?

Ksivusya [100]

Well not getting any of the answers :/ i get 1184

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2 years ago

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25 POINTS AND BRAINLIEST PLEASE HELP ASAP

dexar [7]

M is a midpoint of BC so:

$M=\left(\dfrac{\boxed{2}\boxed{a}+a}{\boxed{2}},\dfrac{\boxed{0}+b}{2}\right)=\left(\dfrac{\boxed{3}\boxed{a}}{\boxed{2}},\dfrac{\boxed{b}}{\boxed{2}}\right)$

Length of MA:

$MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=
\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

Length of NB:

$NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\
\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}$

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3 years ago

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