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3 years ago

Find the perimeter and area of isosceles trapezoid EFGH. A=1/2(b1+b2)•h

Mathematics

1 answer:

vfiekz [6]3 years ago

7 0

We have a height of 4, bottom base of 8, top base of 6, so

A = (1/2)(8+6)(4) = 28

Perimeter, we have sides that are hypotenuses of a right triangle, legs 1 and 4,

s = √(1²+4²) = √17

So perimeter is

P = 8 + 6 + √17 + √17 = 14+ 2√17

Answer: Area 28, perimeter 14 + 2√17

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3 years ago

Find the geometric means in the following sequence.

soldi70 [24.7K]

Answer:

Last choice. d.

Step-by-step explanation:

We are given the first term and the sixth term.

The first term is $a_1=-9$ .

The sixth them is $a_6=a_1r^5=-9216$ .

Let's solve for the common ratio, r.

$-9r^5=-9216$

Divide both sides by -9:

$r^5=1024$

Take the fifth root of both sides:

$r=1024^{\frac{1}{5}}$

$r=4$

So the common ratio is 4.

$a_1=-9$

$a_2=-9(4)=-36$

$a_3=-9(4)^2=-144$

$a_4=-9(4)^3=-576$

$a_5=-9(4)^4=-2304$

$a_6=-9(4)^5=-9216$

8 0

3 years ago

5x – 6y = –29: ( __, 4)

Xelga [282]

Hola, qué tal tu día

7 0

3 years ago

Find the length of the second leg of the triangle round to the nearest hundredth

horsena [70]

Answer:

$\displaystyle \frac{3\sqrt{230}}{5}$

Step-by-step explanation:

Use the Pythagorean Theorem:

$\displaystyle a^2 + b^2 = c^2 \\ \\ 9,7^2 + b^2 = 13,3^2; \sqrt{82,8} = \sqrt{b^2} \\ \\ \frac{3\sqrt{230}}{5}\:[or\:9,0994505329...] = b$

So, you have this:

$\displaystyle 9,1 ≈ b$

I am joyous to assist you at any time.

7 0

2 years ago

Which exponential equation is equivalent to the logarithmic equation below?

julsineya [31]

Answer:

$\implies\boxed{ 10^a = 300 }$

Step-by-step explanation:

We are provided logarithmic equation , which is ,

$\implies log_{10}^{300}= a$

Here we took the base as 10 , since nothing is mentioned about that in Question .

Say if we have a expoteintial equation ,

$\implies a^m = n$

In logarithmic form it is ,

$\implies log_a^n = m$

Similarly our required answer will be ,

$\implies\boxed{ 10^a = 300 }$

3 0

3 years ago