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3 years ago

Find the area of the triangle

Mathematics

1 answer:

Veseljchak [2.6K]3 years ago

7 0

The area of a triangle is equal to bh/2.
The base here is 8 and the height is 3.
8*3=24
24/2=12
The answer is B, 12 yd^2

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Factor 54+27 using the GCF.

e-lub [12.9K]

So, we need the Greatest Common Factor of the two numbers, 54 and 27, right?

first, is one maybe a multiplication of the other?

yes, it is! in fact, 54 is 27*2; 53 is 27 multiplied by 2.

that means that 27 IS the Greatest Common Factor of the two!

3 0

3 years ago

Read 2 more answers

6. A cylindrical oatmeal canister has a diameter of 4 inches and a hight of 10 inches. The manufac-

Arlecino [84]

B. 8 inches, hope this was right and helped:)

6 0

3 years ago

A factory produces 550 bottles of water in 25 minutes. How many bottles of water can it produce in 3 minutes?

vazorg [7]

Answer:

Step-by-step explanation:

step 1: divide 550 by 25

step 2:times your answer by 3

4 0

3 years ago

Can someone help me

Fofino [41]

Step-by-step explanation:

1. Right angled triangle , scalene triangle

2. isosceles triangle, obtuse angled triangle

3. equilateral triangle

4. right angled triangle, isosceles triangle

5. scalene triangle, obtuse angled triangle

6. scalene triangle, acute angled triangle

8 0

2 years ago

Read 2 more answers

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$

Length of curve= $s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})$

Length of curve= $s=32.66$

5 0

3 years ago