All categories

3 years ago

I need help with these two please

Mathematics

2 answers:

jek_recluse [69]3 years ago

4 0

13. Add 2 on both sides. So you get p/4 =2.8 then multiply by 4 on both sides. This cancels out 4 and leave with p= 11.2

damaskus [11]3 years ago

4 0

16 first substract 15 from both sides then you will get s/8 =12 then multiply both sides by 8 to leave s by it self 12 times 8 is 96 so s=96

You might be interested in

If it is 120 miles and you are traveling 40 mph how long will it take you to get to your destination?

Anton [14]

Time = distance/velocity = 120/40 = 3 hours

8 0

3 years ago

Solve the equation for<br> x 3/4(x+21) = 10.5 <br> x =

MrRa [10]

<h3>✍️ Answer:</h3><h3> $x=-\frac{294}{13}$ </h3>

✽✽✽✽

$\frac{x\cdot \:3}{4\left(x+21\right)}=10.5$

$\mathrm{Multiply\:both\:sides\:by\:}4\left(x+21\right)$

$\frac{x\cdot \:3}{4\left(x+21\right)}\cdot \:4\left(x+21\right)=10.5\cdot \:4\left(x+21\right)$

$3x=42\left(x+21\right)$

$3x=42x+882$

$\mathrm{Subtract\:}42x\mathrm{\:from\:both\:sides}$

$3x-42x=42x+882-42x$

$-39x=882$

$\mathrm{Divide\:both\:sides\:by\:}-39$

$\frac{-39x}{-39}=\frac{882}{-39}$

$x=-\frac{294}{13}$

❅❅❅❅❅❅❅❅❅❅

<h3>hope it helps...</h3><h3>have a great day!!</h3>

4 0

2 years ago

Someeeeeoneeee savvveeeee meee

anastassius [24]

The y-intercept is 10 hours and it represents the number of hours it took to empty one tank

3 0

3 years ago

Covert 3.92 into a mixed number in simplest from

Kobotan [32]

First, take three off it, because that will form the whole-number part:
3.92 = 3 + 0.92

Next, read off the place value of the last digit. The 2 is in the 'hundredths' column, which means that 0.92 = 92/100:
3.92 = 3 + 92/100

Finally, simplify 92/100 by dividing top and bottom by 4 to get 23/25. Then, shove it all together:
3.92 = 3 23/25

8 0

3 years ago

Read 2 more answers

Solving fp<br> Algebra 2<br> I do not understand how to solve this, could someone please explain?

Olenka [21]

$f(x) = \sqrt{x + 2}$
What's The Fixed Point? Well, Let's Assume X=0, this is our point.

$f(0) = \sqrt{0 + 2} \\ f(0) = \sqrt{2} \\ this \: is \: our \: fixed \: point \: \\ (0, \sqrt{2} )$
OR

$f(p) = \sqrt{p + 2} \\ \\ or \: \sqrt{x + 2} = 0 \\ ( \sqrt{x + 2} ) ^{2} = {0}^{2} \\ x + 2 = 0 \\ x = - 2$
I'm unaware of the nature of the question, so here are some different ways a fixed point is found based on the merits of the question.

4 0

2 years ago