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3 years ago

In triangle ABC,of angle A=angle B=45°,name the longest side.

Mathematics

1 answer:

serg [7]3 years ago

3 0

The longest side would be C

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Each cube in this figure measures 1 foot on each side.

UNO [17]

Answer:

Your answer would be:
C.) 200 ft³

Step-by-step explanation:

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#TheWizzer

7 0

2 years ago

Read 2 more answers

Write an<br> equation of a line with the given slope and y-intercept.<br> m=-2, b=3

GuDViN [60]

Answer:

y=-2x+3

Step-by-step explanation:

y=mx+b

m=-2,b=3

so, y=-2x+3

hopes this helps

4 0

3 years ago

Can someone help me solve for x and show the work please

Elden [556K]

Answer:

D. -10

Step-by-step explanation:

The sum of a triangle = 180°, therefore,

64° + 63° + x + 63 = 180°

Add like terms

x + 190° = 180°

Subtract 190 from each side

x + 190 - 190 = 180 - 190

x = -10

7 0

3 years ago

X² - 8x - 33=0<br> complete the square

Andreas93 [3]

<h2>Answer:</h2><h2></h2><h2> $\boxed{x_{1}=-3} \\ \\ \boxed{x_{2}=11}$ </h2>

<h2>Explanation:</h2>

Step 1. Write $x^2 - 8x - 33=0$ in the form $x^2+2ax+a^2=0$

$2a=-8\quad :\quad a=-4$

Step 2. Add and subtract $(-4)^2$ to the left side of the equation

$x^2-8x-33+\left(-4\right)^2-\left(-4\right)^2=0 \\ \\ \\ But: \\ \\ x^2+2ax+a^2=\left(x+a\right)^2 \\ \\ x^2-8x+\left(-4\right)^2=\left(x-4\right)^2$

Step 3. Complete square

$\left(x-4\right)^2-33-\left(-4\right)^2=0 \\ \\ \left(x-4\right)^2-49=0 \\ \\ (x-4)=\pm \sqrt{49} \\ \\ x-4=\pm 7 \\ \\ \boxed{x_{1}=-3} \\ \\ \boxed{x_{2}=11}$