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3 years ago

Classify each triangle by its angles and its sides

Mathematics

2 answers:

fenix001 [56]3 years ago

6 0

Answer:

Acute

Step-by-step explanation:

The last angle is 71 which is less than 90

Finger [1]3 years ago

6 0

Answer:

acute triangle

Step-by-step explanation:

The sum of the two angles given is more than 90°, so the remaining angle is less than 90°. All angles are acute angles, so the triangle is an acute triangle.

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Point A is located at (-3, -5)

Vikki [24]

Answer:

(-9, -3)

Step-by-step explanation:

x = -3

y = -5

(x – 6, y + 2)

x = -3 - 6 = -9

y = -5 + 2 = -3

(-9, -3)

8 0

4 years ago

Dont put the file stuff like stop! Help me

grandymaker [24]

5/12lb of dog food in each container.

This seems like a division problem so you would take (5/4) and divide it by 3 to make 3 equal groups meaning 5/12lb would be in each container.

I hope this helps, good luck!

6 0

3 years ago

9) Simplify 2x^5-6^2+4x^4y/2x

KonstantinChe [14]

Answer:

x^4+2x^3y-3x

Step-by-step explanation:

6 0

3 years ago

How many degrees are in the corner of a rectangle

goblinko [34]

90 degrees
Each angle of a rectangle equals 90 degrees.

3 0

3 years ago

Read 2 more answers

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$

Finding c from (1):

$33-56i-7b+4bi+c=0 \\ \\ \\ Substituting \ b: \\ \\ 33-56i-7(14)+4(14)i+c=0 \\ \\ 33-56i-98+56i+c=0 \\ \\ -65+c=0 \\ \\ \boxed{c=65}$

<h2>Learn more:</h2>

Complex conjugate: brainly.com/question/2137496

#LearnWithBrainly

5 0

3 years ago