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NISA [10]
2 years ago
11

Find the area of the triangle ABC

Mathematics
2 answers:
MrRa [10]2 years ago
8 0
6 times 8…………zzzzzzzzzzzzzzzzz
Reil [10]2 years ago
8 0
Answer:

A= 1/2•b•h

A= 1/2•8•6

A=24
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3. At Hector's Torta Shop, customers can order a steak, ham, or chicken sandwich. They can choose
Deffense [45]
I think it’s 96. You just have to multiply 48 (costumers) by 2( one chicken sandwich and one coke)
8 0
3 years ago
HELPPPPPPPPPPPPPP !!!!!!
d1i1m1o1n [39]

Answer:

35 degrees

Step-by-step explanation:

Degrees in a triangle: 180

Assume Angle C = 90 degrees

180-55-90 = 35 degrees

5 0
3 years ago
I need help I’ve been trying to figure this one out could get it
erastovalidia [21]

Answer:

The result is 7.8 * 10^3

Step-by-step explanation:

When we are dividing in scientific notation, divide the numbers in front

7.8/1 = 7.8

Subtract the exponents on the powers of 10

10^(7-4) = 10 ^3

The result is 7.8 * 10^3

3 0
3 years ago
Khan academy<br><br>will mark.​
german

Given:

\theta_1 is located in IV Quadrant.

\sin (\theta_1)=-\dfrac{24}{25}

To find:

The value of \cos \theta_1.

Solution:

We have,

\sin (\theta_1)=-\dfrac{24}{25}

We know that,

\sin^2 \theta+\cos^2\theta=1

So,

\sin^2 (\theta_1)+\cos^2(\theta_1)=1

\left(-\dfrac{24}{25}\right)^2+\cos^2(\theta_1)=1

\cos^2(\theta_1)=1-\dfrac{576}{625}

\cos^2(\theta_1)=\dfrac{625-576}{625}

Taking square root on both sides, we get

\cos (\theta_1)=\pm \sqrt{\dfrac{49}{625}}

\cos (\theta_1)=\pm \dfrac{7}{25}

\theta_1 is located in IV Quadrant. In IV Quadrant the value of cos is positive. So,

\cos (\theta_1)=\dfrac{7}{25}

Therefore, \cos (\theta_1)=\dfrac{7}{25}.

8 0
3 years ago
What steps did you take to bisect an angle?
BaLLatris [955]

Answer:

First, we must have a compass and a rule.

Step-by-step explanation:

1º) Place the tip of the compass in the vertex of the angle and draw a circumference that cuts the two straight lines that form the angle.

2º) Now place the compass at each point of intersection of the circumference of 1º with each line, drawing a circumference of equal radius at each point.

3º) Now we take the point of intersection of the circumferences of equal radius and center in each line that we assemble in point 2, and we join them to the vertex.

4º) In this way we draw the bisecting line to the angle formed by lines 0B and 0A as line 0P.

Have a nice day!

8 0
3 years ago
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