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musickatia [10]
3 years ago
12

What is 9.6+5.42-3.2 squared​

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
7 0

Answer:

139.7124

Step-by-step explanation:

9.6+5.42-3.2 (^2)

11.82^2

= 139.7124

sergey [27]3 years ago
7 0

Answer:

4.78

That’s the answer I guess

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Look at the coordinate graph.
dezoksy [38]

Answer:

B = (5,4), C = (-5, -4), D = (5, -4), A = (-5, 4)

Step-by-step explanation:

look at x axis first (horizontal), then y axis (vertical)

8 0
3 years ago
Bella pays 7 payments of $5 each to a game store.she returns one game receives $20 back.what is the change of the about of money
hichkok12 [17]
15$ would be ya answer ....

6 0
3 years ago
PLEASE HELP I WILL JIVE YOU 100 POINTS NEEDED ASAP
Anastaziya [24]

Answer:

0.3 galloons of water.

Step-by-step explanation:

We know that 1/5 galloons of water fills 2/3 of the bin.

So, we can write the following fraction:

\frac{\frac{1}{5}\text{ galloons}}{\frac{2}{3}\text{ bin}}

We want to find how many galloons will fill the entire bin or 1 whole bin. So, we can write the following proportion:

\frac{\frac{1}{5}\text{ galloons}}{\frac{2}{3}\text{ bin}}=\frac{x\text{ galloons}}{1\text{ bin}}

To find our answer, we need to solve for x.

So, let's cross-multiply. This yields:

\frac{2}{3}x=\frac{1}{5}

Multiply both sides by 3:

3(\frac{2}{3}x)=3(\frac{1}{5})

The left side simplifies. Multiply on the right:

2x=\frac{3}{5}

Now, multiply both sides by 1/2:

\frac{1}{2}(2x)=\frac{1}{2}(\frac{3}{5})

Simplify the left. Multiply on the right:

x=\frac{3}{10}

So, 3/10 of 0.3 of a gallon of water will fill the entire bin.

And we're done!

3 0
3 years ago
Read 2 more answers
Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

5 0
3 years ago
Evaluate 2/3x for x = 3/4. Please show your work.<br> Thank you!!! :)
GenaCL600 [577]
X = 3/4
So substitute x for 3/4 in <span> 2/3x
(3/4) x (2/3)

Now solve
</span>(3/4) x (2/3) = 1/2 or 0.5

5 0
3 years ago
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