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3 years ago

50 to 60 is what percent

Mathematics

1 answer:

Sholpan [36]3 years ago

3 0

83.333333333333%

I hope this helped.

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Which value in the replacement set is a solution to this equation? 2n + 5 = 27 Replacement set: {11, 12, 13} Drag the solution i

Alchen [17]

11 is the answer to your question.

2×11=22

22+5=27

6 0

3 years ago

The temperature in the afternoon was 10 degrees. By 8 pm, the temperature was -5

Ket [755]

Answer:

The temperature had dropped by 15 degrees (-15 if you don't want it positive.)

Step-by-step explanation:

Okay!

This is a subtraction problem.

If it started off at 10, and went down to -5, we can do |-5 - 10| (I am using the two bars because it said change.) And we get 15!

5 0

3 years ago

Read 2 more answers

A number that is prime, between 60-100,1s place is 2 less then 10s place.

Crank

The answer to that would be 97.

Hope this helps! :)

7 0

3 years ago

Will mark Brainlest hellpppp

Anna007 [38]

$h(-3) - h( - 2) = - \frac{5}{ 8} \\ \\$

Step-by-step explanation:

$\boxed{h(x) = \frac{2 {x}^{2} - x + 1}{3x - 2}}$

$h(2) = \frac{2 {(2)}^{2} - (2) + 1}{3(2) - 2}$

$h(2) = \frac{2 {(4)} - 2 + 1}{6 - 2}$

$h(2) = \frac{ 8 - 1}{4}$

$h(2) = \frac{7}{4}$

$h( - 3) = \frac{2 {( - 3)}^{2} - ( - 3) + 1}{3( - 3) - 2} \\ h( - 3) = \frac{2 (9) + 3 + 1}{ - 9- 2} \\ h( - 3) = \frac{18 + 4}{ - 11} \\ h( - 3) = \frac{22}{ - 11} \\ h(-3) = -2$

$h( - 2) = \frac{2 {( - 2)}^{2} - ( - 2) + 1}{3( - 2) - 2} \\ h( - 2) = \frac{2 (4) + 2+ 1}{- 6 - 2} \\ h( - 2) = \frac{8 + 3}{- 8} \\ h( - 2) = \frac{11}{- 8}$

$h(3) - h( - 2) = -2 - \frac{11}{- 8} \\ h(3) - h( - 2) =-2 + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-2}{1}+ \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16}{8} + \frac{11}{ 8} \\ h(3) - h( - 2) = \frac{-16+11}{8} \\ h(3) - h( - 2) = \frac{-5}{8} \\ - \frac{5}{ 8}$

5 0

3 years ago

Pls help me solve this question, also include step by step calculation ya. Thks

sp2606 [1]

$\displaystyle\\Answer:\ x=\sqrt{2}\ \ \ \ \ x=-\sqrt{2}\ \ \ \ \ x=\frac{ln5}{2}$

Step-by-step explanation:

$5x^2-x^2e^{2x}+2e^{2x}=10$

$Let:\ x^2=t \ \ \ \ \ e^{2x}=v\\Hence,\\5t-tv+2v^2=10\\5t-tv+2v^2-10=10-10\\5t-tv-10+2v^2=0\\t(5-v)-2(5-v)=0\\(5-v)(t-2)=0\\5-v=0\\5-v+v=0+v\\5=v\\5=e^{2x}$

Prologarithmize both parts of the equation:

$ln5=lne^{2x}\\ln5=2x*lne\\ln5=2x*1\\ln5=2x\\$

Divide both parts of the equation by 2:

$\displaystyle\\x=\frac{ln5}{2}$

$t-2=0\\t-2+2=0+2\\t=2\\x^2=2\\x=\sqrt{2} \\x=-\sqrt{2}$

6 0

1 year ago