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2 years ago

Find the area of the triangle with A= 60° b=11 feet and c=8 feet

Mathematics

1 answer:

Rudik [331]2 years ago

5 0

Answer:

5280

Step-by-step explanation:

if u times them all up first u get 60 times 11 witch = 660 and then i times 660 and 8 to get 5280

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Willis bought a gallon of paint.He painted a wall that is 9 feet high and 10 feet wide.Then he used he rest of the paint to pain

Otrada [13]

First you need to find the surface area of the wall that was painted, and then add the additional area in the hallway that was painted.

The wall that was painted was 9 feet by 10 feet, or 90 square feet. 90 square feet plus the additional 46 square feet in the hallway is 136 square feet, which is the total area that the paint covered.

Therefore the solution is 136 square feet.

7 0

3 years ago

Read 2 more answers

Look at the table below.

GenaCL600 [577]

Answer:04

Step-by-step explanation:

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2 years ago

How can you tell if a equation is extraneous

geniusboy [140]

Do you mean an extraneous solution? I will tell you what I think. To tell if a solution is extraneous you need to go back to the original problem and check if there actually is a solution or if the solution is actually right. If not then it is extraneous.

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3 years ago

Mathematical induction, prove the following two statements are true

adelina 88 [10]

Prove:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}$
____________________________________________

Base Step: For n=1:
$n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1$
and
$4-\dfrac{n+2}{2^{n-1}}=4-3=1$
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}$
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}$

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
$=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}$

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
$1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}$

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
$=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}$

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
$=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}$

Distribute the -2 and combine the fractions together,
$=4+\dfrac{-2k-4+(k+1)}{2^{k}}$

Combine like-terms,
$=4+\dfrac{-k-3}{2^{k}}$

pull the negative back out,
$=4-\dfrac{k+3}{2^{k}}$

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
$=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}$

3 0

2 years ago

PLEASSEEE HELPPPP ME EITH THIS!!

Hunter-Best [27]

Answer:

PQ = 98.21

Step-by-step explanation:

PQ/23 = 47/11

PQ × 11

23 × 47

11PQ = 1081

11PQ/11 = 1081/11

PQ = 98.21

7 0

2 years ago