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3 years ago

Marked it 30 points so please help also idk how to show my work for first question

Mathematics

2 answers:

Darina [25.2K]3 years ago

7 0

Answer:

(9,0) and (-4,0)

Step-by-step explanation:

First, move everything to the left to make the whole equation equal to 0.

So we get x^2-5x-36=0

Then we can factor by finding the two numbers that multiply to -36 and add to -5.

Those two numbers are -9 and 4.

So we have (x-9)(x+4)=0.

So the two roots are 9 and -4.

To verify this on a graphing calculator, plug in the equation and look at the x intercepts.

If they match up with (9,0) and (-4,0) you have your answer.

igomit [66]3 years ago

7 0

Answer:

x=-4 and 9 are the 2 solutions

Step-by-step explanation:

Let's use factoring

first, get one side 0 by subtracting 36 from both sides

$x^2-5x-36=0$

now factor, think, what 2 numbers multiply to get -36 and add to get -5?

-9 and 4

so it factor into

$(x-9)(x+4)=0$

set each to 0 and solve for x

x-9=0

x=9

x+4=0

x=-4

so the 2 solutions are x=-4 and 9

To verify, you want to see where $y=x^2-5x$ and $y=36$ intersect

If they intersect at x=-4 and x=9, then your answer is correct

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Point R on the coordinate grid below shows the location of a car in a parking lot. A second car needs to park 6 units above R to

Over [174]

Answer:

$R' = (-4 , 4)$

Step-by-step explanation:

Given

See attachment for the grid coordinate

From the attachment, we have that:

$R = (-4,-2)$

Required

Determine the coordinates of the second truck

Represent the position with R'

From the question, we understand the position of R' is 6 above R

When a point (a,b) is moved n point above, the new point becomes (a, b + n)

So:

$R = (-4,-2)$

When moved 6 points above becomes:

$R' = (-4 , -2 + 6)$

$R' = (-4 , 4)$

Hence, (-4,4) answers the question

7 0

4 years ago

Write ratio in simplest form . 5km : 450 cm

liubo4ka [24]

5km = 5000m = 50000cm
thus, 50000:450 then simpify it. the final answer would be 1000:9

5 0

4 years ago

Please help with these 2 math problems. ( Multiple choice) im begging. Please don't just do it for points I really need help

Brrunno [24]

The first one and A and the second one is c

3 0

3 years ago

1 х Given g(x)=

yarga [219]

(a) Since $g(x)=\sqrt[3]{x}$ and $h(x) = \frac1{x^3}$ , we have

$(g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x$

We're given that

$(f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}$

but we can rewrite this as

$\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}$

(bear in mind that we can only do this so long as x ≠ 0) so it follows that

$f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}$

(b) On its own, we may be tempted to conclude that the domain of $(f\circ g\circ h)(x) = \frac1{1+x}$ is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.

$g(x) = \sqrt[3]{x}$ is defined for all x - no issue here.

$h(x) = \frac1{x^3}$ is defined for all x ≠ 0. Then $(g\circ h)(x) = \frac1x$ also has a domain of x ≠ 0.

$f(x) = \frac1{1+x}$ is defined for all x ≠ -1, but

$(f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}$

is undefined not only at x = -1, but also at x = 0. So the domain of $(f\circ g\circ h)(x)$ is

$\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}$

7 0

3 years ago

Multiply. 29 · 22 · 24 PLZ HELP

kati45 [8]

Answer:

2^15

Step-by-step explanation: You add the Exponents together

8 0

3 years ago

Read 2 more answers