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kodGreya [7K]
3 years ago
6

Help please and that you 13-16

Mathematics
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

See explanation.

Step-by-step explanation:

13. (4*5)/2 = 10

14. 5*7=35 7-3=4 8-5=3 (4*3)/2 = 6 35+6 = 41

15. 3*4 = 12

16. (4*4)/2 = 8 8*4=32 4*2=8 16*8=128 128-32 = 96

Let me know if you want me to better explain how I solved any of these.

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Use the distributive property to remove the parentheses.
sergij07 [2.7K]

Answer:

3+2x

Step-by-step explanation:

Distribute the equation out.

1/4(12+8x)

(1/4*12)+(1/4*8x)

Multiple those out to get a new equation

3+2x

You cannot add those together because of the variable. Theres no way of knowing what x equals so this is as far as you can go therefor making the answer 3+2x

7 0
3 years ago
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15 less than quadruple a number w​
vichka [17]

Answer:

4w-15

Step-by-step explanation:

For expressions that have "more than/less than" , that number goes AFTER the "whatever a number" statement because that's just how they explain it. I don't remember the exact reason why, but that is how my professor made me remember it, and yes, the order WILL affect the expression since that's a specific topic and I suppose you're being tested on how to make algebraic expressions based on descriptions.

8 0
3 years ago
Which of the following is a function equivalent to f(x) = 5(1.9)3x?
Sophie [7]
B. f(x)=5(6.86x) this the correct anwser
6 0
3 years ago
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What is the reflection across the y-axis?
Allisa [31]

Answer:

c

Step-by-step explanation:

what ever it is you do the same on the other side

3 0
1 year ago
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Determine
mihalych1998 [28]

Answer:

=4.+2.

Step-by-step explanation:

<u>Linear Combination Of Vectors </u>

One vector \vec b is a linear combination of \vec a_1 and \vec a_2 if there are two scalars x_1, x_2 such as

\vec b=x_1\vec a_1+x_2\vec a_2

In our case, all the vectors are given in R^3 but there are only two possible components for the linear combination. This indicates that only two conditions can be used to determine both scalars, and the other condition must be satisfied once the scalars are found.

We have

\vec a_1=,\ \vec a2=,\ \vec b=

We set the equation

=x_1.+x_2.

Multiplying both scalars by the vectors

=+

Equating each coordinate, we get

4x_1-4x_2=8

5x_1+3x_2=26

-4x_1+3x_2=-10

Adding the first and the third equations:

-x_2=-2

x_2=2

Replacing in the first equation

4x_1-4(2)=8

4x_1=8+8

x_1=4

We must test if those values make the second equation become an identity

5(4)+3(2)=20+6=26

The second equation complies with the values of x_1 and x_2, so the solution is

=4.+2.

8 0
3 years ago
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