Given: f(x)=4x-9 and g(x)= 2x-3. What is the value of (f+g)(0)=?A.-9
B.-10
C.-11
D.-12

Answers

Answer 1

Answer:

Answer:

D. -12

Step-by-step explanation:

if f=4x-9, and g=2x-3, and the value is (f+g), then we have to add the equations. This is equal to 6x-12. And then, in '(f+g)(0)', it is asking what is the y-value at x=0. Plug in 0 into the equation, and the y-value equals -12.

-6

Step-by-step explanation:

The given translation is

$T_(-3,-8) (x,y)$

Which means that the original figure will be moved 3 units to the right and 8 units downwards.

Remember, when it comes to translations, when we subtract units to $x$ that means the figure will be moved rightwards. And if we subtract units from $y$ , that means the figure will be moved downwards.

So, the original figure has as vertex B(1,2). If we apply the transformation to its vertical cordinate $y=2$ , we would have

$y'=2-8=-6$

Therefore, the right answer is the third choice -6.

Which choice is equivalent to the expression below when y 0?√Y^+√9Y^3-3Y√Y

A.Y√10Y-3Y√Y
B.-2Y√11Y
C.Y√Y
D.√10Y^3-3Y√Y

Answers

$y√(y)+√(9y^3)-3y√(y)=y√(y)+\sqrt9\cdot√(y^2\cdot y)-3y√(y)\n\n=y√(y)+3y√(y)-3y√(y)=y√(y)\n\nAnswer:C.$

The square of a number plus the number is 42 . Find the number (s).

Answers

Answer: s = 6 or s = -7

Step-by-step explanation:

Our task is to find the number. But first, let's write the equation:

The square of a number s plus the number is 42.

The square of s = $\sf{s^2}$

Plus s = $\sf{s^2+s}$

this equals 42: $\sf{s^2+s=42}$

Now, let's solve for s. Subtract 42 from both sides, and set the right-hand side equal to 0.

$\sf{s^2+s-42=0}$

Now, let's factor it.

Think of two numbers that:

multiply to -42
add up to 1

These numbers are 6 and -7.

So, the factored expression is :

(s + 6)(s - 7) = 0

Either s + 6 is 0 or s - 7 = 0.

We have two little equations that we can solve for s:

s + 6 = 0 s - 7 = 0

s = 0 - 6 s = 0 + 7

s = -6 s = 7

Final answer:

To find the number(s) that satisfy the equation x^2 + x = 42, you can factorize the quadratic equation (x + 7)(x - 6) = 0 and solve for x. The solutions are x = -7 and x = 6.

Explanation:

To find the number(s) that satisfies the equation, let's represent the number as 'x'. According to the question, the square of the number plus the number is equal to 42. This can be written as x^2 + x = 42. Rearranging the equation, we get x^2 + x - 42 = 0. To solve this quadratic equation, we can factorize it or use the quadratic formula.

Factoring the quadratic equation, we find (x + 7)(x - 6) = 0. Setting each factor equal to zero, we get x + 7 = 0 and x - 6 = 0. Solving these equations, we find x = -7 and x = 6.

Therefore, the number(s) that satisfy the equation are x = -7 and x = 6.

Learn more about Finding solutions to a quadratic equation here:

brainly.com/question/34125066

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If there is 360 grams of radioactive material with a half-life of 8 hours, how much of the radioactive material will be left after 32 hours and is the radioactive decay modeled by a linear function or an exponential function?A. 22.5 grams; linear

B. 22.5 grams; exponential

C.45 grams; linear

D. 45 grams; exponential

Answers

B.) 22.5 GRAMS, EXPONENTIAL

32 hours ÷ 8 hours = 4

The radioactive material will half 4 times.

360 x (1/2)⁴ = 22.50

360 x 1/2 = 180
180 x 1/2 = 90
90 x 1/2 = 45
45 x 1/2 = 22.50

Answer:

B. 22.5 grams; exponential .

Step-by-step explanation:

We have been given that there is 360 grams of radioactive material with a half-life of 8 hours.

As amount of radioactive material remains 1/2 of the amount after each 8 hours, therefore, our function will be an exponential decay function.

We will use half-life formula to solve our given problem.

$y=a*((1)/(2))^{(t)/(b)}$ , where,

$a=\text{Initial value}$ ,

$t=\text{Time}$ ,

$b=\text{Half life}$ .

Let us substitute a=360 and b=8 in half life formula to get half life function for our given radioactive material.

$y=360*((1)/(2))^{(t)/(8)}$ , where y represents remaining amount of radioactive material after t hours.

Therefore, the function $y=360*((1)/(2))^{(t)/(8)}$ gives the half-life of our given radioactive material.

Let us substitute t=32 in our half life function to find the amount of material left after 32 hours.

$y=360*((1)/(2))^{(32)/(8)}$

$y=360*((1)/(2))^(4)$

$y=360*(1^4)/(2^4)$

$y=360*(1)/(16)$

$y=22.5$

Therefore, the radioactive material will be left 22.5 grams after 32 hours and the radioactive decay is modeled by an exponential function and option B is the correct choice.

Which values, when placed in the box, would result in a system of equations with no solution? Check all that apply.y = –2x + 4

6x + 3y =

A: –12

B: –4

C: 0

D: 4

E: 12

Answers

For this case we have the following system of equations:

Rewriting equation 1 we have:

Therefore, the equivalent system is:

The system will have no solution, if we write equation 2 as a linear combination of equation 1.

Therefore, since both lines have the same slope, they are parallel.

Parallel lines do not intersect when they have different cut points.

Therefore, there is no solution for:

-12, -4, 0, 4

The system has inifinites solutions for:

This is because the lines intersect at all points in the domain.

Answer:

The values, when placed in the box, would result in a system of equations with no solution are:

A: -12

B: -4

C: 0

D: 4

Answer:

Thus, option (a), (b) , (c) , (d) are correct.

The system will have no solution for all values except 12.

Step-by-step explanation:

Given a system of equation y = –2x + 4 and 6x + 3y = ?

We have to check for which value the '?' would result in a system of equations with no solution.

Consider a system of equation $a_1x+b_1y+c_1=0 \n\na_2x+b_2y+c_2=0$

For the system to have no solution the condition is,

$(a_1)/(a_2)=(b_1)/(b_2)\neq (c_1)/(c_2)$

For the given system of equation ,

Let unknown quantity be v.

y = –2x + 4 ⇒ 2x +y - 4 = 0

and 6x + 3y = v ⇒ 6x + 3y - v =0

On comparing, we get,

$a_1=2 , b_1=1,c_1=-4\n\n\a_2=6,b_2=3,c_3=-v$

Substitute the values in condition for no solution , we get ,

$(2)/(6)=(1)/(3)\neq (-4)/(-v)$

Consider second and third ratio, we get,

$(1)/(3)\neq (-4)/(-v)$

Solve for v , we get,

$(1)/(3)\neq (-4)/(-v) \n\n \Rightarrow v \neq 12$

Thus, for all values v except v = 12

The system will have no solution

at v = 12 the system will have infinite many solution.

Thus, option (a), (b) , (c) , (d) are correct.

Is #1 correct and I need help on #2 THANKS :)

Answers

Answer 1 = I believe you are correct cause when doing area , remembering you times the different area of the sides

Answer 2 = I believe the answer is C (2,0009) cause remember adding when doing volume but there are few steps ..
. i have done the work for this and hoping my work helped the answer i got help you! comment below if you need anything , thanks , have a nice day!!

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