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3 years ago

Original:1.5new:2.5show work please

Mathematics

1 answer:

andre [41]3 years ago

4 0

Answer:

The question is kinda incorrect but if it is.... The answer should be 1

Step-by-step explanation:

Original :1.5 + 1

New: 2.5

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Calculate the sum. 10+ (-42) O A. 52 O B. -32 O C. 32 D. -52

kotykmax [81]

Answer:

Step-by-step explanation:

10 + (-42)

10 - 42

-32

8 0

3 years ago

A client pays for a moving company to move 25 boxes. If the moving company charges $12.50 per box and a base fee of $199.95 to e

rewona [7]

Answer:

512.45

Step-by-step explanation:

312+199.95=512.45

3 0

3 years ago

The price of a pair of jeans is $25. If there is a 15% discount, what is the sale price of the jeans?

dusya [7]

Answer:

$29.41

Step-by-step explanation:

You know the discount price is 25 and the original is unknown so call it x. The sale is 15% off and that means that the jeans are 85% of what they used to be since 100-15 is 85. Then you plug it into a formula of 25/x = 85/100 then you cross multiply to get 2500=85x and when you solve for x and round you get $29.41

4 0

3 years ago

Lim x--> 0 (e^x(sinx)(tax))/x^2

Tom [10]

Make use of the known limit,

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

We have

$\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)$

since $\tan x=\dfrac{\sin x}{\cos x}$ , and the limit of a product is the same as the product of limits.

$\dfrac{e^x}{\cos x}$ is continuous at $x=0$ , and $\dfrac{e^0}{\cos 0}=1$ . The remaining limit is also 1, since

$\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1$

so the overall limit is 1.

4 0

3 years ago

Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

5 0

3 years ago

Original:1.5new:2.5​show work please

Original:1.5new:2.5show work please