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2 years ago

How do I solve numbers 8 and 9?

Mathematics

1 answer:

Tresset [83]2 years ago

7 0

8: 20
9: 90
I am so sorry if this is wrong. I am not good at this type of things

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The sides of an equilateral triangle are 8 units long.what is the length of the altitude of the triangle?

dedylja [7]

Answer:

Equilateral triangle implies each angle is 60 degree.

And any median is the altitude.

So, NT perpendicular LM, where T is the midpoint of [ML];

We use the Pythagorean Theorem in right triangle NTL ( NT = h; NL = a;

LT = a / 2 );

h² = a² - (a/2)² = a² - a²/4 = 4a²/4 - a²/4 = 3a²/4 ;

It follows that

h = √(3a²/4) = a√3/2 ≈ 0,866*a;

But, a = 8;

h ≈ 0,866 * 8 ≈ 6,928;

Step-by-step explanation:

6 0

4 years ago

20 points.Cynthia has a bag of jellybeans. There are two red jellybeans, one yellow jellybean, and two black jellybeans in her b

solmaris [256]

Answer:

2/5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Solve the given inequality :

tigry1 [53]

Answer:

x < -5 or x = 1 or 2 < x < 3 or x > 3

Step-by-step explanation:

Given rational inequality:

$\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0$

$\textsf{Factor }(x^2-5x+6):$

$\implies x^2-2x-3x+6$

$\implies x(x-2)-3(x-2)$

$\implies (x-3)(x-2)$

Therefore:

$\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0$

Find the roots by solving f(x) = 0 (set the numerator to zero):

$\implies (x-1)^2(x-2)^3=0$

$\implies (x-1)^2=0\implies x=1$

$\implies (x-2)^3=0 \implies x=2$

Find the restrictions by solving f(x) = undefined (set the denominator to zero):

$\implies (x-3)^2(x-2)^2(x+5)=0$

$\implies (x-3)^2=0 \implies x=3$

$\implies (x-2)^2=0 \implies x=2$

$\implies (x+5)=0 \implies x=-5$

Create a sign chart, using closed dots for the roots and open dots for the restrictions (see attached).

Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.

Test values: -6, 0, 1.5, 2.5, 4

For each test value, determine if the function is positive or negative:

$f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+$

$f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-$

$f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

$f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+$

Record the results on the sign chart for each region (see attached).

As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).

Therefore, the solution set is:

x < -5 or x = 1 or 2 < x < 3 or x > 3

As interval notation:

$(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)$

4 0

2 years ago

Someone helppppp me on this math problem

KiRa [710]

Answer:

whats the problem? ill do my best to help

Step-by-step explanation:

Hope this helps! Plz give brainliest! Also, plz sub to Kgirl633 on yt.

7 0

3 years ago

A total of 511 tickets were sold for the school play. They were either adult tickets or student tickets. There were 61 more stud

Marina86 [1]

Answer:

255

Step-by-step explanation:

EXPLANATION We are looking for the number of adult tickets sold. number of adult tickets sold =x We are given that there were 61 more student tickets sold than adult tickets. number of student tickets sold =+x61 The total number of tickets sold was 511.

5 0

3 years ago