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3 years ago

Find the graphed solution to the inequality. 3(x-9) < -18

Mathematics

1 answer:

Ket [755]3 years ago

6 0

Answer:

$x$

Step-by-step explanation:

$3\left(x-9\right)$

<u>Divide both sides by 3</u>

$\frac{3\left(x-9\right)}{3}$

$x-9$

<u>Add -9 to both sides:</u>

$x-9+9$

$x$

_______________________

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PLEASE HELP ME WHOEVER HELPS I WILL MARK BRAINLEST :)

julsineya [31]

Remember the shortcut way for graphing quadratic equations

A quadratic function has graph as parabola
Hence on both sides of vertex the parabola is symmetric and axis of symmetry is vertex x values .
y on both sides for -x and +x is same

y=2(x-3)²-1

Vertex

(h,k)=(3,-1)

As a is positive parabola facing upwards

Find y for same x distance from vertex

I took 3-1=2 and 3+1=4

f(2)=2(2-3)²-1=2(-1)²-1=2-1=1
f(4)=2(4-3)²-1=1

Now plot vertex and these two points (2,1) and (4,1) on graph then draw a parabola by freehand

y=(x-2)(x+4)
y=x²+4x-2x-8
y=x²+2x-8

Convert to vertex form

y=x²+2x+1-9
y=(x+1)²-9

Vertex at (-1,-9)

Same take two equidistant x values

Let's take -1-1=-2 and -1+1 =0

f(-2)=(-2+1)²-9=1-9=-8
f(0)=(1)²-9=-8

Put (-1,-9),(-2,-8),(0,-8) on graph and draw a freehand parabola

#3.

Yes it can be verified by finding the coordinate theoretically on putting them on function then can be verified through putting them on graph whether they matches or not

8 0

2 years ago

Suppose that the probability a seed will germinate is 85%. What is the probability that 7 of these seeds will germinate when 10

Viktor [21]

Total number of outcomes of 7 from 10 = 10*9*8 / 3*2*1 = 120
probability of each outcoe is 0.85^7* (0.15)^(3) = 0.0010819

Required probability = 120 * 0.0010819 = 0.1298

6 0

4 years ago

Read 2 more answers

The equation p(t) = 1.e represents a

zvonat [6]

Answer:

(b), (d) and (e)

Step-by-step explanation:

Given

$p(t) = 1 * e^t$

See attachment for $y = p(t)$

Required

Select true statements from the given options

(a) $\ln(30)$ = days the bacteria reaches 30000

We have:

$p(t) = 1 * e^t$

In this case:

$t = \ln(30)$ and $p(t) = 30000$

So, we have:

$30000 = 1 * e^{\ln(30)}$

$30000 = e^{\ln(30)}$

Using a calculator, we have:

$e^{\ln(30)} = 30$

So:

$30000 = 30$

The above equation is false.

(a) is not true

(b) The graph shows that $\ln(20) \approx 3$

We have:

$p(t) = 1 * e^t$

Let t = 3

So;

$p(3) = 1 * e^3$

From the graph, $p(3) = 20$

So:

$20 = 1 * e^3$

$20 = e^3$

Take natural logarithm of both sides

$\ln(20) = \ln(e^3)$

This gives:

$\ln(20) = 3$

(b) is true

(c) $\ln(t) = y$ is the logarithm form of $y = e^t$

We have:

$y = e^t$

Take natural logarithm of both sides

$\ln(y) = \ln(e^t)$

This gives:

$\ln(y) = t$

$\ln(y) = t$ $\ne$ $\ln(t) = y$

(c) is false

(d) $e^4 > 50$ and $\ln(50) < 4$

From the graph, we have:

$e^4 = 54$ --- rough readings

This implies that:

$e^4 > 50$ is true

Because $54 > 50$

Take natural logarithm of both sides

$\ln(54) > \ln(50)$

Rewrite as:

$\ln(50) < \ln(54)$

We have:

$e^4 = 54$

Take natural logarithm of both sides

$\ln(e^4) = \ln(54)$

$4 = \ln(54)$

$\ln(54) = 4$

Substitute $\ln(54) = 4$ in $\ln(50) < \ln(54)$

$\ln(50) < 4$

(d) is true

(e) The graph shows that $10 \approx \ln(2.3)$

We have:

$p(t) = 1 * e^t$

Let t = 2.3

So;

$p(2.3) = 1 * e^{2.3}$

From the graph,

$p(2.3) = 10$ ---- rough readings

So:

$10 = 1 * e^{2.3}$

$10 = e^{2.3}$

Take natural logarithm of both sides

$\ln(10) = \ln(e^{2.3})$

This gives:

$\ln(10) = 2.3$

(e) is true

4 0

3 years ago

Multiply (-5)^3 (raised to the third power)

gayaneshka [121]

Answer:

- 125

Step-by-step explanation:

$(-5)^{3}$

= - 5 × - 5 × - 5

= 25 × - 5

= - 125

8 0

3 years ago

Read 2 more answers

I need help on this one

tia_tia [17]

You need to come up with an array of answers for x and y that will be greater than 0.
x+4y>0. example if x is -1 and y is 1 will this be correct.
-1+(4*1)>0, -1+4>0. 3>0? Yes
if x is 1 and y is -1 then the equation becomes 1+(-4)>0 -3 is not>0 so it's not in the array
x could be -3 and y 1
You will plot a line to show all the answers.
i

8 0

4 years ago