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Aleonysh [2.5K]
2 years ago
13

Please help me with the answer

Mathematics
2 answers:
liraira [26]2 years ago
5 0

Answer:

-18

Step-by-step explanation:

-15+6=-9

the absolute value of -9 is 9

and 9*-2=-18

wariber [46]2 years ago
3 0

Answer:

-18

Step-by-step explanation:

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katen-ka-za [31]
A.)Rachel should have 6 second class tramps. Because if you do the equation 6+6+6 you get 18 (the number of first class stamps she has)
2.)Add 6 to 18 to get your answer :)
6 0
3 years ago
2/5 of 40 in fractional parts
Romashka [77]
Dividing 40 into fifths...
40 ÷ 5 = 8

Since 1/5 of 40 is 8, 2/5 of 40 is <u>16</u>.
7 0
3 years ago
Read 2 more answers
Which scale factors produce a contraction under a dilation of the original image? Select each correct answer. ​−6​ ​−0.5​ 0.5 6
Step2247 [10]

A contraction would make the image smaller.


A negative number reflects the image and changes the size based on the number.


To make an image smaller the factor needs to be less than 1

so -0.5 and 0.5 would make the image smaller since 0.5 is less than 1.

7 0
3 years ago
Read 2 more answers
Chris pays a fee if her balance falls below $10 on the statement data. Prior to the statement date, her balance was -$3.46. Then
alisha [4.7K]

Answer:

The solution of the inequality required for the situation is d  ≥ $13.46.

Step-by-step explanation:

i) the minimum deposit is $10.

ii) the balance before the statement is $-3.46

iii) a deposit d is made so that the fee did not have to be made

iv) therefore d + (-3.46)  ≥  10

                    d - 3.46 ≥ 10

    therefore d  ≥ 10 + 3.46

 therefore d  ≥ $13.46

3 0
3 years ago
How do I do this? please detail steps.
vladimir2022 [97]
Define
{x} =   \left[\begin{array}{ccc}x_{1}\\x_{2}\end{array}\right]

Then
x₁ = cos(t) x₁(0) + sin(t) x₂(0)
x₂ = -sin(t) x₁(0) + cos(t) x₂(0)

Differentiate to obtain
x₁' = -sin(t) x₁(0) + cos(t) x₂(0)
x₂' = -cos(t) x₁(0) - sin(t) x₂(0)

That is,
\dot{x} =   \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right] x(0)

Note that
\left[\begin{array}{ccc}0&1\\-1&09\end{array}\right]   \left[\begin{array}{ccc}cos(t)&sin(t)\\-sin(t)&cos(t)\end{array}\right] =  \left[\begin{array}{ccc}-sin(t)&cos(t)\\-cos(t)&-sin(t)\end{array}\right]

Therefore
x(t) =   \left[\begin{array}{ccc}0&1\\-1&0\end{array}\right] x(t)

7 0
3 years ago
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