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3 years ago

12

HELP ME‍♂️ILL MARK BRAINLIESTTTT‼️‼️Find the value of x that makes the quadrilateral a parallelogram.

1 answer:

AysviL [449]3 years ago

3 0

Answer:

x = 23 and y = 20

Step-by-step explanation:

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During 30 minutes on the treadmill, LaToya burned 20 calories per minute. What is the rate of change of the function that repres

Genrish500 [490]

The answer I believe is 10 calories per minute because 20 x30 =600 I did that because she burned 20 calories per minute so with that being said there’s 60 minutes is the maximum minutes any time can go to so you do 600 divided by 60 which equals 10 which means she burns 10 calories per minute hope this is right

3 0

3 years ago

Which equation has the solution set x = (23) ?

Svet_ta [14]

Answer:

If you mean 2 and 3 it would be (x-2)(x-3)=0

Step-by-step explanation:

(x-2)=0

x-2=0 move the two on the other side to get positive

x=2

(x-3)=0

x-3=0 move the three on the other side to get positive

x=3

4 0

3 years ago

Find 7x+y when x=4 and y=5. *

Alexeev081 [22]

Answer:

7x+y, when x+4 and y+5

7(4) + 5=

28+5=

33

Step-by-step explanation:

Hope this helps. :)

7 0

3 years ago

Is this correct? Bowmanian

BartSMP [9]

Yes it is correct!!!!

7 0

3 years ago

Read 2 more answers

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

Viktor [21]

Answer:

Therefore $k= \frac{ln2 }{18}$ , A=184

Step-by-step explanation:

Given function is

$T(t)=230 -e^{-kt}$

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

$138=230 -Ae^{-k\times 18}$

$\Rightarrow -Ae^{-18k}=138-230$

$\Rightarrow Ae^{-18k}=92$ .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

$184=230-Ae^{-k\times 36}$

$\Rightarrow Ae^{-36k}=230-184$

$\Rightarrow Ae^{-36k}=46$ .......(2)

Dividing (2) by (1)

$\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}$

$\Rightarrow e^{-18k}=\frac{46}{92}$

Taking ln both sides

$ln e^{-18k}=ln\frac{46}{92}$

$\Rightarrow -18k =ln (\frac12)$

$\Rightarrow -18k= ln1-ln2$

$\Rightarrow k= \frac{ln2 }{18}$

Putting the value k in equation (1)

$Ae^{-18\frac{ln2}{18}}=92$

$\Rightarrow A e^{ln2^{-1}}=92$

$\Rightarrow A.2^{-1}=92$

$\Rightarrow \frac{A}{2}=92$

$\Rightarrow A= 92 \times 2$

⇒A= 184.

Therefore $k= \frac{ln2 }{18}$ , A=184

7 0

3 years ago