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3 years ago

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

Read 2 more answers

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