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3 years ago

For model 5, the tip of the blade travels 0.18 miles what is the approximate speed

Mathematics

1 answer:

Whitepunk [10]3 years ago

3 0

0.9 is the answer to your question

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three sets of the sum of a number and four are added to the sum of seven times the same number and thirteen

almond37 [142]

The algebraic expression for given statement is: 10x + 25 or 3(x + 4) + (7x + 13)

Solution:

Given the statement:

Three sets of a sum of a number and four are added to the sum of seven times the same number and thirteen

Let us first understand the given statement,

Let the number be "x"

" sum of a number and four" means x + 4

"Three sets of a sum of a number and four" translated to 3(x + 4)

"sum of seven times the same number and thirteen" means 7x + 13

Thus the algebraic expression for given statement is:

$3(x+4)+(7x + 13)$

Using distributive property in above expression

$a(b+c) = ab + bc$

Therefore,

$3(x+4)+(7x + 13) = 3x + 12 + 7x + 13$

Combine the like terms

$3(x+4)+(7x + 13) = 10x+25$

Thus the required expression for given statement is: 10x + 25 or 3(x + 4) + (7x + 13)

7 0

3 years ago

SIMPLE INTEREST

cupoosta [38]

The answer would be 5 years. I=PxRxT.

3 0

3 years ago

Read 2 more answers

The circumference of a circle is 60 kilometers. What is its radius?

lisabon 2012 [21]

Answer: 9.5493

Step-by-step explanation: circumference = 2pi*r

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3 years ago

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Element X decays radioactively with a half-life of eight minutes. If there are 800 g of element X, how long, to the nearest 10th

Harman [31]

Answer: 41.5 min

Step-by-step explanation:

This problem can be solved with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=22 g$ is the final amount of the radioactive element

$A_{o}=800 g$ is the initial amount of the radioactive element

$t$ is the time elapsed

$h=8 min$ is the half life of the radioactive element

So, we need to substitute the given values and find $t$ from (1):

$22 g=(800 g) 2^{\frac{-t}{8 min}}$ (2)

$\frac{22 g}{800 g}=2^{\frac{-t}{8 min}}$ (3)

$\frac{11}{400}=2^{\frac{-t}{8 min}}$ (4)

Applying natural logarithm in both sides:

$ln(\frac{11}{400})=ln(2^{\frac{-t}{8 min}})$ (5)

$-3.593=-\frac{t}{8 min}ln(2)$ (6)

Clearing $t$ :

$t=41.46 min \approx 41.5 min$ This is the time elapsed

7 0

3 years ago

A 44 ft ladder leans against a building so that the top of the ladder touches the building at a point 27 ft above the ground. To

Veseljchak [2.6K]

Answer is option 4. working out attached

hope it helps :)

4 0

4 years ago