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horsena [70]
3 years ago
12

I beg you to help me .

Mathematics
1 answer:
choli [55]3 years ago
7 0

Answer:

16+3=y

y=19

Step-by-step explanation:

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I know life can get hard but don't give up no matter what happens, because you never get anything out of quitting, but you get t
ss7ja [257]

Answer: Nice words my guy

Step-by-step explanation:

8 0
3 years ago
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The factor (x + 2) occurs in the numerator and denominator. There will be a hole at x = 2.
navik [9.2K]

Answer:

false

Step-by-step explanation:

it suppose to be -2 if you equate it to Zero

6 0
3 years ago
If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
4 years ago
What is the answer to this question please explain (picture included)
Yakvenalex [24]
Easy peasy
just use PEMDAS and some exonential laws

(x^{m})(x^{n})=x^{m+n}
(x^{m})^{n}=x^{mn}
(ab)^{m}=(a^{m})(b^{m})

also another is x^{\frac{m}{n}}=\sqrt[n]{x^{m}}

so

[3(2a)^{\frac{3}{2}}]^{2}=
(3^{2})((2a)^{\frac{3}{2}}^{2})=
(9)((2a)^{\frac{3}{2}}^{2})=
(9)((2a)^{\frac{6}{2})=
(9)((2a)^{3}) =
(9)(8a^{3}) =
72a^{3}

3 0
3 years ago
Square root of 214 show work thanks
Nataliya [291]
√214= 14.6287388383278

Calculator maybe??
6 0
3 years ago
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