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3 years ago

Can someone please help me pleaaaase ill give brainliest :'((((

Mathematics

2 answers:

NeX [460]3 years ago

4 0

Answer:

$2^{x}$ + 8x - 5

Step-by-step explanation:

(f + g)(x) = f(x) + g(x), thus

f(x) + g(x)

= $2^{x}$ + 5x + 3x - 5 ← collect like terms

= $2^{x}$ + 8x - 5

horsena [70]3 years ago

4 0

Answer: 2^x+8x-5 i did the work on paper, You're welcome!

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sally has 20 coins in her piggy bannk, all dimes and quarters. the total amount of money is $3.05. how many of each coin does sh

VMariaS [17]

D=dimes
q=quarters

d + q = 20
.10d + .25q = 3.05

d=-q + 20

.10(-q + 20) +.25q = 3.05
-.10q + 2 + .25 = 3.05
.15q + 2 = 3.05
.15q + 2 - 2 = 3.05 - 2
.15q = 1.05
.15q/.15 = 1.05/.15
q = 7

d + 7 = 20
d + 7 - 7 = 20 - 7
d = 13

Sally has 7 quarters and 13 dimes.
Hope this helps!

5 0

3 years ago

what do you do before adding the equation so that one variable will be eliminated when you dd them 2x-4y=5 and 6x-3y=10

Naddika [18.5K]

We have a system of equations 2x - 4y = 5 and 6x - 3y = 10.
We want to eliminate one variable when we add up the equations.

If we want to eliminate the x variable, we need to multiply the top, bottom or both of the equations with a number that when the equations are added together will eliminate the x variable.

Multiply the top equation by -3.

-3(2x - 4y) = 5 * -3
-6x + 12y = -15

Now when we add the two equations together, the x's will be eliminated.

-6x + 12y = -15
6x + 6y = 10
18y = -5

The x's were eliminated.

6 0

4 years ago

A machine has four components, A, B, C, and D, set up in such a manner that all four parts must work for the machine to work pro

olchik [2.2K]

Answer:

0.756

Step-by-step explanation:

It is given that a machine has four components, A, B, C, and D.

$P(A)=P(B)=0.93, P(C)=0.95,P(D)=0.92$

If these components set up in such a manner that all four parts must work for the machine to work properly.

We need to find the probability that the machine works properly. It means we have to find the value of $P(A\cap B\cap C\cap D)$ .

If two events X and Y are independent, then

$P(X\cap Y)=P(X)\times P(Y)$

Assume the probability of one part working does not depend on the functionality of any of the other parts.

$P(A\cap B\cap C\cap D)=P(A)\times P(B)\times P(C)\times P(D)$

Substitute the given values.

$P(A\cap B\cap C\cap D)=0.93\times 0.93\times 0.95\times 0.92$

$P(A\cap B\cap C\cap D)=0.7559226$

$P(A\cap B\cap C\cap D)\approx 0.756$

Therefore, the probability that the machine works properly is 0.756.

7 0

3 years ago

Gnoma [55]

Answer: Third option.

Step-by-step explanation:

These are some transformations for a function f(x):

If $f(x)-k$ , then the function is shifted down "k" units.

If $f(x-k)$ , then the function is shifted right "k" units.

Knowing this we can describe the transformation from the graph of the function $f(x) = x^2$ to the graph of the function $f(x) = (x - 3)^2-1$ . This is: