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3 years ago

Will give brainliest

Mathematics

1 answer:

klio [65]3 years ago

8 0

Answer:

Area of the parallelogram =base×height

area of parallelogram =3cm×2cm

area of parallelogram = 6cm²

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Expressions equal to 10^5

Drupady [299]

Answer:

10 x 10 x 10 x 10 x 10, 20 x 20 x 10, 30 x 20

6 0

3 years ago

Can u help sold this

solong [7]

Answer:

Step-by-step explanation:

To calculate the slope or gradient we use this formula:

Slope = y2-y1/x2-x1

(-3,2) = (x1, y1)

(4,2) = (x2, y2)

Slope = 2-2/4-(-3) = 0

Answer from Gauthmath

4 0

3 years ago

HELP What is the area of a circle with a radius of 13 feet? IT IS NOT 530.93 ft

NeTakaya

Answer:

530.66

Step-by-step explanation:

to find the area you need to do pi times the radius square and to find the radius squared you just times it by its self so 13 times 13 which is 169 and pi is 3.14

so now you just do 3.14x169=530.66

8 0

3 years ago

Which expression has the same value as 4÷25?

Hatshy [7]

Answer:

We need the choices, since there are plenty equations with the same value.

Step-by-step explanation:

4/25=0.16

Whichever equation equals 0.16 is your answer.

6 0

3 years ago

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