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Bad White [126]

3 years ago

10

Sue wrote a doubles fact. It has a sum less than 10 and greater than 4. The addends are each less than 5. What fact might she ha

ve written?

1 answer:

harina [27]3 years ago

6 0

2×4=8 because 2 and 4 would be the addends and their less than 5 and it equal an answer between 4 and 10

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Which two values of x are roots of the polynomial below? x^2-3x+5

Gemiola [76]

Given:

Polynomial $x^2-3x+5$

To find:

The values of x.

Solution:

$x^2-3x+5=0$

Quadratic equation formula:

$$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here $a=1, b=-3, c=5$ .

Substitute the values in the formula, we get

$$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \cdot 1 \cdot 5}}{2 \cdot 1}$

$$x=\frac{3 \pm \sqrt{9-20}}{2}$

$$x=\frac{3 \pm \sqrt{-11}}{2}$

$$x=\frac{3 - \sqrt{-11}}{2} \ \text{and} \ x=\frac{3 + \sqrt{-11}}{2}$

Option E and option F are the roots of the polynomial.

The values of x are $x=\frac{3 - \sqrt{-11}}{2} \ \text{and} \ x=\frac{3 + \sqrt{-11}}{2}$ .

7 0

3 years ago

If 3(r+300)=6, then what would be the value of r+300-2?

saveliy_v [14]

Solve for r in the first equation:

3(r+300) = 6

Use the distributive property:

3r + 900 = 6

Subtract 900 from both sides:

3r = -894

Divide both sides by 3:

r = -894 / 3

r = -298

Now you have r, replace r in the second equation and solve:

r +300 -2 =

-298 + 300 - 2 = 0

The answer is 0.

6 0

3 years ago

Silas just bought a used car for $5000. It loses value at a rate of 30% per year. The value of the car in n years is modeled by

Fantom [35]

Answer:

$1200.50

plug 5 for n

5000(.7)^4

5000*.2401

=1200.5

8 0

2 years ago

Hey I need this answered i have no clue what to put pls help

Molodets [167]

Answer:

b is correct for Maria but c is correct for the boy

4 0

2 years ago

There are 5 red balls, 4 blue balls, 6 yellow balls and 10 green balls in a box,

lys-0071 [83]

<h2 /><h2>Here we go ~ </h2>

According to given information there are :

5 red balls

4 blue balls

6 yellow balls

10 green balls

<h3>1. what is the probability that the ball chosen is red ?</h3>

-

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{total \: red \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{5}{25}$

$\qquad \sf \dashrightarrow \: p(red) = \dfrac{1}{5}$

<h3>2. what is the probability that the ball chosen is blue ?</h3>

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{total \: blue \: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(blue) = \dfrac{4}{25}$

<h3>3. what is the probability that the ball chosen is yellow ?</h3>

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{total \: yellow\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(yellow) = \dfrac{6}{25}$

<h3>4. what is the probability that the ball chosen is green ?</h3>

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{total \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{10}{25}$

$\qquad \sf \dashrightarrow \: p(green) = \dfrac{2}{5}$

<h3>5. what is the probability that the ball chosen is not green ?</h3>

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{total \: non \: green\: balls}{total \: balls}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{5 + 4 + 6}{5 + 4 + 6 + 10}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{15}{25}$

$\qquad \sf \dashrightarrow \: p(not \: green) = \dfrac{3}{5}$

3 0

2 years ago