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3 years ago

Can someone please help me solve this?

Mathematics

2 answers:

RUDIKE [14]3 years ago

5 0

I think the answer would be 9x:)

mezya [45]3 years ago

4 0

Sorry this is quite hard but will you answer any question from me and here is the answer 9x feet

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Which statement is correct?

Alenkinab [10]

Answer:

The domain of the graph is all real numbers less than or equal to 0.

Hope this helps! :D

3 0

3 years ago

Find the following:

Butoxors [25]

Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say $\displaystyle \lim_{x\rightarrow a}f(x)=L$ , if f(x) approaches L as x approaches 'a'.

(a)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\$

$\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

(b)

$\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\$

$\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8$

3 0

3 years ago

Find the approximate side length of a square game board with an area of 105 in2.

aleksley [76]

Answer:

Rounded to the nearest inch: 10

Rounded to the nearest tenth: 10.2

Rounded to the nearest hundredth: 10.25

Step-by-step explanation:

To find the area of a square, we multiply the side length by itself, or in other words, SQUARE the side length!

$A=s^2$

We are given the area, so plugging that into our equation looks like this:

$105=s^2$

To eliminate the exponent, take the square root of both sides of the equation!

$\sqrt{105}=s$

From here, there are a few paths to take, depending on what your teacher is looking for. I will evaluate the approximate of the square root by finding the nearest perfect squares.

The nearest perfect square less than √105 is √100, and the nearest perfect square greater than √105 is √121.

105 is closer to 100 than to 121, so √105 is closer to √100.

√100 is equal to 10, so the √105 is approximately 10 inches!

Using a calculator to check:

$\sqrt{105}=10.247$ inches

8 0

1 year ago

Ron needed to buy a new jacket for his upcoming ski trip. The ticket price of the jacket was $199.00. After the sales tax was ad

Levart [38]

Answer: 5%

Step-by-step explanation:

Given that:

Ticket price of jacket = $199

Final price after addition of sales tax = $208.95

Percentage sales tax on the jacket =?

Final price after addition of sales tax = price of jacket + amount of sales tax paid

208.95 = 199 + amount of sales tax paid

amount of sales tax paid = 208.95 - 199

amount of sales tax paid = $9.95

Let Percentage of sales tax = x

(x / 100) * 199 = 9.95

199x / 100 = 9.95

199x = 995

x = 5

Hence percent sales tax paid = 5%

8 0

3 years ago

Decrease £120 by 15%

Irina-Kira [14]

To decrease this amount by 15% we must first find out what is 15% percent.

We can do that by multiplying the decimal of 15% by 120

15% = 0.15

0.15 x 120 = 18

So 18 is 15% of 120.

Wait! that's not the answer. We are asked to decrease by 15%

So 120 - 18 = 102

Your answer is £102

6 0

3 years ago

Can someone please help me solve this?​

Can someone please help me solve this?