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3 years ago

Point M is the midpoint of AB . AM= 3x+3, and AB=8x−6

Mathematics

2 answers:

Studentka2010 [4]3 years ago

6 0

<h2>Answer:</h2>

The length of AM is:

21 units.

<h2>Step-by-step explanation:</h2>

Point M is the midpoint of AB.

AM= 3x+3, and AB=8x−6

Since, the midpoint divides the line segment into two equal parts i.e. it bisects the line segment.

Hence, we have:

AM+MB=AB

Also, AM=MB

Hence, we have:

$3x+3+3x+3=8x-6$

on combining the like terms in the left hand side of the equation we have:

$3x+3x+3+3=8x-6\\\\6x+6=8x-6$

Now, on subtracting both side of the equation by 6x we have:

$6=8x-6-6x\\\\6=8x-6x-6\\\\6=2x-6$

on adding 6 on both side of the equation we have:

$6+6=2x\\\\12=2x\\\\2x=12\\\\x=\dfrac{12}{2}\\\\x=6$

Hence, we have:

$AM=3\times 6+3\\\\AM=21\ \text{units}$

algol133 years ago

3 0

AM= Half of AB

or, 3x+3=(8x-6)/2
or, 6x+6=8x-6
or, 2x=12
Therefore,x=6

so,AM=3*6+3=21

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The percentage of students who spend less than 4 hours a week on screens which reported a grade average below 80 is 50%

Given, we have the report representing children spending time online, on social media or watching T.V, which has there effect on there grades as above or below 80.

By the question we know that the number of students who spends less than 4 hours a week on screen is:

9 + 9

= 18

And among the 18 students there are 9 students whose grade average is below 80.

So the percentage of the students who spends 4 hours a week on screen is:

= 9 ÷ 18 × 100%

= 100/2

= 50%

hence 50% students spend less than 4 hours a week on screens.

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1 year ago

. An individual wishes to invest $5000 over the next year in two types of investment: Investment A yields 5%, and investment B y

tresset_1 [31]

Answer:

The amount invested at Investment A must be greater than or equal to $2,750

The amount invested at Investment B must be less than or equal to $2,250

Step-by-step explanation:

Let

x -----> the amount invested at Investment A yields 5%

y -----> the amount invested at Investment B yields 8%

we know that

$x+y=5,000$

$x=5,000-y$ -----> equation A

$x \geq 0.25(5,000)$

$x \geq \$1,250$ -----> inequality B

$y \leq 0.50(5,000)$

$y \leq \$2,250$ -----> inequality C

$x \geq \frac{1}{2}y$ -----> inequality D

Substitute equation A in the inequality D and solve for y

$5,000-y \geq \frac{1}{2}y$

Multiply by 2 both sides

$10,000-2y \geq y$

Multiply by -1 both sides

$-10,000+2y \leq -y$

Adds y both sides

$-10,000+2y+y \leq 0$

$-10,000+3y \leq 0$

adds 10,000 both sides

$3y \leq 10,000$

Divide by 3 both sides

$y \leq \$3,333.33$ -----> inequality E

therefore

<em>Solve for y</em>

we have

$y \leq \$2,250$ -----> inequality C

$y \leq \$3,333.33$ -----> inequality E

The solution of inequality C and inequality E is

$y \leq \$2,250$

For y=2,250

x=5,000-y ----> x=5,000-2,250=2,750

$x \geq \$2,750$ -----> inequality F

<em>Solve for x</em>

we have

$x \geq \$1,250$ -----> inequality B

$x \geq \$2,750$ -----> inequality F

The solution of inequality B and inequality F is

$x \geq \$2,750$

therefore

The amount invested at Investment A must be greater than or equal to $2,750

The amount invested at Investment B must be less than or equal to $2,250

7 0

3 years ago

There are 380 students at Cove Elementary School. The students voted on

Mrrafil [7]

yes because you need to do this then you need to do that and plus yeah you right

6 0

3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

miss Akunina [59]

Answer:

a. 0.343

b. 0.657

c. 0.189

d. 0.216

e. 0.353

Step-by-step explanation:

We use the combination formula of probability distribution to solve the question.

Where P(x=r) = nCr * p^r * q^n-r

Where n = number of trials = 3 vehicles

r = desired outcome of trial which varies.

p = probability of success = 70% =0.7

q = probability of failure = 1-p = 0.3

a. Probability that all 3 vehicles passed = P(X=3)

= 3C3 * 0.7^3 * 0.3^0 = 1 * 0.343 * 1

= 0.343.

b. Probability that at least one fails = 1 - (probability that none failed)

And probability that none failed = probability that all 3 vehicles passed.

Hence Probability that at least one fails = 1 - (probability that all 3 vehicles passe)

= 1 - 0.343

= 0.657

c.) probability that exactly one pass= P(X=1)

= 3C1 * 0.7¹ * 0.3² = 3 * 0.7 * 0.09

= 0.189

d.) probability that at most One of the vehicles passed = probability that none passed + probability of one passed.

Probability that none of the vehicles passed = P(X=0)

= 3C0 * 0.7^0 * 0.3^3 = 1*1*0.027

=0.027

Probability that one passed as calculated earlier = 0.189

Hence probability that at most one vehicle passed = 0.189 + 0.027 = 0.216

e.) Probability that all three Vehicles pass given that at least one pass = (probability of all three vehicles passes) / (probability that at least one passes)

Probability that at least one pass = 1 - probability that none passed.

= 1 - 0.027

= 0.973

Hence,

Probability that all three Vehicles passed given that at least one passed = 0.343/0.973

= 0.3525 = 0.353 (3.d.p)

7 0

3 years ago

2. What's the area of a circle with radius 18 units? A. 36? units2 B. 324? units2 C. 18? units2 D. 9? units2

e-lub [12.9K]

c. 18 is the answer you get

3 0

3 years ago

Read 2 more answers