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3 years ago

15

I need help ill give out brainliest please don’t answer if you don’t know thank you

1 answer:

ratelena [41]3 years ago

7 0

Step-by-step explanation:

I think it's right answer is c

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Look at the image below

Alexandra [31]

that's what I got, you can confirm if it's correct or not... I'm not sure

have a nice dayʘ‿ʘ

6 0

2 years ago

Learning goal: to understand the rules for computing dot products. let vectors a=(2,1,−4), b=(−3,0,1), and c=(−1,−1,2). part a -

miv72 [106K]

For this case we have the following vectors:
$a = (2,1, -4)

b = (- 3,0,1)

c = (- 1, -1,2)$
The dot product of two vectors is a scalar.
The point product consists of multiplying component by component and then adding the result of the multiplication of each component.
For the product point of the vectors a and b we have:
$a.b = (2,1, -4). (- 3,0,1)

a.b = (2) (- 3) + (1) (0) + (-4) (1)

a.b = - 6 + 0 - 4

a.b = - 10$
Answer:
The product point of the vectors a and b is:
$a.b = - 10$

3 0

3 years ago

Which unit would you use to measure the amount of water in a swimming pool?

ycow [4]

The unit u have to use to measure the amount of water in a swimming pool is in Liters :)

8 0

2 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

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

3 years ago

#3 and 5 please<br> I don’t understand how to get them

denpristay [2]

Answer:

im bad with this kinda thing, just guess and start praying, goodluck

Step-by-step explanation:

6 0

2 years ago