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2 years ago

Please answer I really need help due at 11:50

Mathematics

1 answer:

klemol [59]2 years ago

5 0

Answer:

|-3| = 3

|3| = 3

Step-by-step explanation:

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Which graph represents-3x - 4y ≤ 2?

den301095 [7]

Answer:

Step-by-step explanation:

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3 years ago

Given: △ABC, AB=5 square root 2 m∠A=45°, m∠C=30° Find: BC and AC

hichkok12 [17]

Answer:

Part a) $BC=10\ units$

Part b) $AC=13.66\ units$

Step-by-step explanation:

step 1

Find the length side BC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{BC}{sin(A)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$C=30^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{BC}{sin(45^o)}$

solve for BC

$BC=\frac{5\sqrt{2}}{sin(30^o)}(sin(45^o))$

$BC=10\ units$

step 2

Find the measure of angle B

we know that

The sum of the interior angles in a triangle must be equal to 180 degrees

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$45^o+m\angle B+30^o=180^o$

$75^o+m\angle B=180^o$

$m\angle B=180^o-75^o$

$m\angle B=105^o$

step 3

Find the length side AC

Applying the law of sines

we know that

$\frac{AB}{sin(C)}=\frac{AC}{sin(B)}$

we have

$AB=5\sqrt{2}\ units$

$A=45^o$

$B=105^o$

substitute

$\frac{5\sqrt{2}}{sin(30^o)}=\frac{AC}{sin(105^o)}$

solve for AC

$AC=\frac{5\sqrt{2}}{sin(30^o)}(sin(105^o))$

$AC=13.66\ units$

4 0

3 years ago

A right triangle has legs that are 18 centimeters and 27 centimeters long.

kvv77 [185]

Here you go, you have to use Pythagorean’s theorem

6 0

3 years ago

Find the length (distance) of a segment with endpoints (5,-2) and (9,4). (Must show your

stepladder [879]

Answer:

$7,2 ≈ D$

Step-by-step explanation:

We need to use the Distance Formula:

$\sqrt{(-x_{1} + x_{2})^{2} + (-y_{1} + y_{2})^{2}} = D \\ \\ \sqrt{(-5 + 9)^{2} + (2 + 4)^{2}} = D \\ \\ \sqrt{{4}^{2} + {6}^{2}} = D \\ \\ \sqrt{16 + 36} = D \\ \\ 2\sqrt{13} = D \\ \\ 7,2 ≈ D$

** Since we are talking about distance, we ONLY want the NON-NEGATIVE root.

I am joyous to assist you anytime.

6 0

3 years ago

the parking lot of the mall has a capacity of 231 cars . O n sunday 3/11 of the parking lot was occupied with cars .how many add

Blababa [14]

8/11 of the lot was unoccupied. Thus, the # of add'l cars that could be parked there would be (8/11)(231 cars) = 168 (answer).

5 0

3 years ago

Read 2 more answers