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4 years ago

Please help on this equation: 3(4v-1)-8v=17. Thank You

1 answer:

6 0

First distribute the 3 into 4v-1 which will result in:
12v-3-8v=17

then add/subtract your v values, so 12v-8v, which leaves:
4v-3=17

then add 3 to both sides, which cancels it out on the left side, leaving:
4v=17+3 simplified to 4v=20

next divide both sides by 4, which cancels out the 4 on the left side, leaving:
v=20/4 simplified to v=5

Good luck!

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HELP please urgent!! - PLEASE CLICK, NEED HELP -

Makovka662 [10]

Answer:

$y=3x^2-12x-135$

Step-by-step explanation:

The standard form of a quadratic is $y=ax^2+bx+c$

We will use the x and y values from each of our 3 points to find a, b, and c. Filling in the x and y values from each point:

First point (-5, 0):

$0=a(-5)^2+b(-5)+c$ and

0 = 25a - 5b + c

Second point (9, 0):

$0=a(9)^2+b(9)+c$ and

0 = 81a + 9b + c

Third point (8, -39):

$-39=a(8)^2+b(8)+c$ and

-39 = 64a + 8b + c

Use the elimination method of solving systems on the first 2 equations to eliminate the c. Multiply the first equation by -1 to get:

-25a + 5b - c = 0

81a + 9b + c = 0

When the c's cancel out you're left with

56a + 14b = 0

Now use the second and third equations and elimination to get rid of the c's. Multiply the second equation by -1 to get:

-81a - 9b - c = 0

64a + 8b + c = -39

When the c's cancel out you're left with

-17a - 1b = -39

Between those 2 bolded equations, eliminate the b's. Do this by multiplying the second of the 2 by 14 to get:

56a + 14b = 0

-238a - 14b = -546

When the b's cancel out you're left with

-182a = -546 and

a = 3

Use this value of a to back substitute to find b:

56a + 14b = 0 so 56(3) + 14b = 0 gives you

168 + 14b = 0 and 14b = -168 so

b = -12

Now back sub in a and b to find c:

0 = 25a - 5b + c gives you

0 = 75+ 60 + c so

0 = 135 + c and

c = -135

Put that all together into the standard form equation to get

$y=3x^2-12x-135$

6 0

4 years ago

Read 2 more answers

Can you plz tell me how to work it out plz I need it but today

Tatiana [17]

First we gather all the information we have about the amount of miles that Carolyn runs:
Monday, Wednesday, Friday: 4 1/2 miles
Tuesday, Saturday: 2 3/4 miles
And the question is the amount of miles that she runs in 4 weeks.
So lets begin by calculating the amount of miles she runs in one week, then we just multiply by 4 and we'll have the final answer.
So, we have to add 4 1/2 three times (Monday, Wed, Friday), and 2 3/4 two times (Tuesday, Sat):
1 week Miles = 4 1/2 + 4 1/2 + 4 1/2 + 2 3/4 + 2 3/4
lets add the whole parts and the fraction parts apart:
1 week Miles = (4 + 4 + 4 + 2 + 2) + (1/2 + 1/2 + 1/2 + 3/4 + 3/4)
1 week <span>Miles = (16) + (3/2 + 6/4)
</span>we can reduce the last fraction:
1 week <span>Miles = (16) + (3/2 + 3/2)
</span>1 week <span>Miles = (16) + (6/2)
</span>1 week <span>Miles = (16) + (3)
</span>1 week <span>Miles = 19
</span>therefore Carolyn runs 19 miles each week, so for 4 weeks we have to add 19 four times or multiply it by 4, is the same:
4 weeks <span>Miles = 4*19
</span>4 weeks <span>Miles = 76
hence Carolyn runs 76 miles in 4 weeks.</span>

4 0

3 years ago

Read 2 more answers

Consider the following vector function. R(t) = 9 2 t, e9t, e−9t (a) find the unit tangent and unit normal vectors t(t) and n(t)

garik1379 [7]

The unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

<h3>What is vector?</h3>

It is defined as the quantity that has magnitude as well as direction also the vector always follows the sum triangle law.

We have vectored function:

$\rm R(t) = (9\sqrt{2t}, e^{9t}, e^{-9t})$

Find its derivative:

$\rm R'(t) = (9\sqrt{2}, 9e^{9t}, -9e^{-9t})$

Now its magnitude:

$\rm |R'(t) |= \sqrt{(9\sqrt{2})^2+ (9e^{9t})^2+ (-9e^{-9t})^2}$

After simplifying:

$\rm R'(t) = 9 \dfrac{e^{18t}+1}{e^{9t}}$

Now the unit tangent is:

$\rm T(u) = \dfrac{R'(t)}{|R'(t)|}$

After dividing and simplifying, we get:

$\rm T(u) = \dfrac{1}{e^{18t}+1} (\sqrt{2}e^{9t}, e^{18t}, -1)$

Now, finding the derivative of T(u), we get:

$\rm T'(u) = \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t}), 18e^{18t}, 18e^{18t})$

Now finding its magnitude:

$\rm |T'(u) |= \dfrac{1}{(e^{18t}+1)^2} (9\sqrt{2}e^{9t}(1-e^{18t})^2+ (18e^{18t})^2+( 18e^{18t})^2)$

After simplifying, we get:

$\rm |T'(u)|= \dfrac{9\sqrt{2}e^{9t}}{e^{18t}+1}$

Now for the normal vector:

Divide T'(u) and |T'(u)|

We get:

$\rm N(t) = \dfrac{1}{e^{18t}+1} ( 1-e^{18t}, \sqrt{2}e^{9t}, \sqrt{2}e^{9t})$

Thus, the unit tangent vector is T(u) and the unit normal vector is N(t) if the vector function. R(t) is equal to 9 2 t, e9t, e−9t.

Learn more about the vector here:

brainly.com/question/8607618

#SPJ4

3 0

2 years ago

Use functional notations to describe the function displayed to the right.

LekaFEV [45]

Examining the table the functional notation is in the second option which is f ( x ) = x^2

<h3>How to determine the functional notations of the table</h3>

The table represents functions of input variables and the output variables.

The functional notation represents the relationship between the input and out variables

comparing the values and the function