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Mumz [18]
3 years ago
6

Evaluate the expression, given functions f and h: f(x) = 3x − 1, h(x) = −2x2 + 3x − 7.

Mathematics
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

Your answer is 27

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A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that co
White raven [17]

Answer:

The value of x would be \frac{3}{5}y or \frac{3}{8}z

Step-by-step explanation:

Given,

x tons of 10% gravel G mixture is mixed with y ton of 2% gravel G to obtain  5% gravel mixture,

Thus, gravel G in x ton + gravel in y ton = gravel G in the resultant mixture,          

⇒ 0.1x + 0.02y = 0.05(x+y),

0.1x - 0.05x = 0.05y - 0.02y

0.05x = 0.03y

\implies x = \frac{0.03}{0.05}y=\frac{3}{5}y----(1)

Now, resultant mixture is z ton,

i.e. z=x+y

\implies y = z - x

From equation (1),

x=\frac{3}{5}(z-x) = \frac{3}{5}z-\frac{3}{5}x

x+\frac{3}{5}x=\frac{3}{5}z

\frac{8}{5}x=\frac{3}{5}z

x=\frac{3}{8}z

3 0
3 years ago
There are 57 students in the orchestra and twice that number in the band. There are 18 boys and 28 girls in the choir. If each s
Klio2033 [76]
There are 217 students total
7 0
3 years ago
Read 2 more answers
Multi-step K is the midpoint of JL, JL=4x-2, and JK=7. Find x, KL, and JL
ddd [48]
If K is midpoint of JL then JK = 0.5JL

JL = 4x - 2; JK = 7

The equation:
0.5(4x - 2) = 7
2x - 1 = 7      |add 1 to both sides
2x = 8        |divide both sides by 2
<u>x = 4</u>

<u>JL</u> = 4(4) - 2 = 16 - 2 = <u>14</u>

<u>KL</u> = JK =<u> 7</u>
7 0
3 years ago
HELLPPPPPPPP ASAPPPPPPPP PLEASEEEEE
marin [14]

Answer:

B=fh

Step-by-step explanation:

I think?

Basically divided by b to get h over b and then times by h to get b=fh (you want to be get on its own)

4 0
3 years ago
Read 2 more answers
PLEASE HELP QWQ AsAp with these 4 questions
den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]  and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)] which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)] which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)] which gives us:

-126+48-90-[-280+54-36] which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right] = \left[\begin{array}{c}-5\\-2\\\end{array}\right] and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]  (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right] Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right] and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

y = -11

(It works, I checked it)

7 0
3 years ago
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