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3 years ago

What is the formula to the area of a triangle?

Mathematics

2 answers:

lyudmila [28]3 years ago

8 0

Answer:

A=bh/2

Step-by-step explanation:

For a right triangle with base b and a height h, its area is A=bh/2

Lorico [155]3 years ago

7 0

Answer:

(Base x height) x 1/2

Step-by-step explanation:

We do this because a triangle is emulating a rectangle with Base x Height. However, a triangle's shape is less conventional than a standard square or rectangle, thus the division by 2.

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Anna has let out 50 meters of kite string when she observes that her kite is directly above emily. If anna is 35 meters from emi

zloy xaker [14]

If you draw a picture you can see that the kite, Anna and Emily form a right triangle.

so you can use the pythagorean theorem to find the answer.

$35 ^{2} + b^{2} = 50^{2}$

1,225 + b^{2} = 2,500

b^{2} = 1,275

b = 35.7 meters

3 0

3 years ago

Read 2 more answers

What would my GPA be? Here are my grades.

marta [7]

That is approximately, in order, a C, a C, a B, a D, an A, a B, and an A.

In grade points that is a 2, a 2, a 3, a 1, a 4, a 3, and a 4.

The average of those numbers is about 2.7, so you have a 2.7.

You can raise that by bringing up the D as it is an outlier here.

4 0

3 years ago

If (-1, y) lies on the graph of y=2^2x then y=

zubka84 [21]

Answer:

y is 1/4 when x is -1 if I did interpret your equation right. Please see that I did.

I believe the equation to be $y=2^{2x}$ . If you meant $y=2^2x$ , please let me know.

Thanks kindly.

Step-by-step explanation:

If (-1,y) lies on the graph of $y=2^{2x}$ , then by substitution we have:

$y=2^{2\cdot -1}$ .

We just need to simplify to determine y:

$y=2^{2 \cdot -1}$

Multiply the 2 and -1:

$y=2^{-2}$

Use reciprocal rule for exponents to get rid of the negative:

$y=\frac{1}{2^2}$

$2^2$ just means $2(2)$ :

$y=\frac{1}{2(2)}$

$y=\frac{1}{4}$ .

8 0

2 years ago

How to evaluate the limit

anzhelika [568]

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$ , which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$