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uranmaximum [27]

3 years ago

13

The graph of the function g(x)= x^2+3x-4 is shifted 5 units to the left

1 answer:

omeli [17]3 years ago

5 0

<h2><em><u>HEYA</u></em></h2>

<h3>THE MID TERM SPLITTING OF THE EQUATION IS :</h3>

${x}^{2} + 3x - 4$

${x}^{2} + 4x - x - 4$

$x(x + 4) - 1(x + 4)$

$x = -4 \: and \: 1$

$x = 1$

now shifted to 5 units left

therefore,

x=-4

hope it helps you mate thanks for the question and if possible please mark it as brainliest

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How do I solve this equation?

frosja888 [35]

Answer:

x=-3/11

Step-by-step explanation:

(x+3)/5 +2x=0

Multiply both sides by 5 to get rid of the fraction

5*( x+3)/5 +5*2x=0*5

x+3 + 10x =0

Combine like terms

11x +3 = 0

Subtract 3 from both sides

11x+3-3 =0-3

11x = -3

Divide by 11

11x/11 = -3/11

x = -3/11

4 0

2 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$

= $(\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5$

= $(\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]$

= $(\frac{3}{4^4})[1+\frac{1}{16}]$

= $(\frac{3}{256})[\frac{17}{16}]$

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0

2 years ago

A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t

VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field= $1500000$ square ft

Area of rectangular field= $l\times b$

Area of rectangular field= $x\time y$

$1500000=xy$

$y=\frac{1500000}{x}$

Fencing used ,P(x)= $x+x+y+y+y=2x+3y$

Substitute the value of y

P(x)= $2x+3(\frac{1500000}{x})$

$P(x)=2x+\frac{4500000}{x}$

Differentiate w.r.t x

$P'(x)=2-\frac{4500000}{x^2}$

Using formula: $\frac{dx^n}{dx}=nx^{n-1}$

$P'(x)=0$

$2-\frac{4500000}{x^2}=0$

$\frac{4500000}{x^2}=2$

$x^2=\frac{4500000}{2}=2250000$

$x=\sqrt{2250000}=1500$

It is always positive because length is always positive.

Again differentiate w.r.t x

$P''(x)=\frac{9000000}{x^3}$

Substitute x=1500

$P''(1500)=\frac{9000000}{(1500)^3}>0$

Hence, fencing is minimum at x=1 500

Substitute x=1 500

$y=\frac{1500000}{1500}=1000$

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used= $2(1500)+3(1000)=6000 ft$

Hence, the shortest length of fence that the rancher can used=6000 ft

3 0

3 years ago

How to re arrange 6x+2y=3 to y= mx+b form

Alex

Answer:

y=-3x+3/2

Step-by-step explanation:

move 6x to the other side

2y=-6x+3

then divide y

y=-3x+3/2

8 0

2 years ago

PLS ANSWER THIS ASAP

Alona [7]

It is the first answer choice

5 0

2 years ago

Read 2 more answers