All categories

2 years ago

The painting shown on the right has an area of 300 in^2. what is the value of x?

Mathematics

1 answer:

Nataly [62]2 years ago

4 0

Answer: X = 10.20240940...

Step-by-step explanation:

x(2x + 9) = 2x^2 + 9x

2x^2 + 9x = 300

- 300 ON BOTH SIDES

2x^2 + 9x - 300 = 0

SOLVE USING THE QUADRATIC FORMULA

x = -b +/- all root (b)^2 - 4(a)(c) All over 2(a)

When all the values are plugged in:

When using "+" in the equation you should get:

x = 10.20240940…

When using "-" in the equation you should get:

x = −14.70240940…

Now.. you CANNOT have a negative length, so you cross of the negative value leaving you one value for x which is 10.20240940...

YOUR ANSWER IS: x = 10.20240940...

You might be interested in

A fossil was analyzed and determined to have a carbon-14 level that is 80 % that of living organisms. the half-life of c-14 is 5

liq [111]

Answer:

The fossil is 1860 years old.

Step-by-step explanation:

The equation for the amount of fossil has the following format:

$Q(t) = Q(0)e^{rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount and r is the rate of change.

Half-life of c-14 is 5730 years.

This means that $Q(5730) = 0.5Q(0)$

$Q(t) = Q(0)e^{rt}$

$0.5Q(0) = Q(0)e^{5730r}$

$e^{5730r} = 0.5$

$\ln{e^{5730r}} = \ln{0.5}$

$5730r = \ln{0.5}$

$r = \frac{\ln{0.5}}{5730}$

$r = -0.00012$

$Q(t) = Q(0)e^{-0.00012t}$

How old is the fossil?

This is t for which

$Q(t) = 0.8Q(0)$

$Q(t) = Q(0)e^{-0.00012t}$

$0.8Q(0) = Q(0)e^{-0.00012t}$

$e^{-0.00012t} = 0.8$

$\ln{e^{-0.00012t}} = \ln{0.8}$

$-0.00012t = \ln{0.8}$

$t = -\frac{\ln{0.8}}{-0.00012}$

$t = 1860$

The fossil is 1860 years old.

4 0

3 years ago

How many centimeters are in 740.2

Tpy6a [65]

74.02 centimeters are in 740.2.

6 0

3 years ago

Read 2 more answers

You can make chili powder from fresh chilis. If 1 chili makes 1 tablespoon of chili powder, how many chilis should you buy?

Tresset [83]

It depends on how much chili powder you want to make. Probably just like one or two

6 0

3 years ago

James states that quadrilateral

mr_godi [17]

Using the distance formula,

$AD=\sqrt{(-1-3)^{2}+(-3-(-4))^{2}}=\sqrt{17}\\\\BC=\sqrt{(5-9)^{2}+(1-0)^{2}}=\sqrt{17} \\\\\therefore \overline{AD} \cong \overline{BC}$

$AB=\sqrt{(-1-5)^{2}+(-3-1)^{2}}=\sqrt{52}=2\sqrt{13}\\\\CD=\sqrt{(9-3)^{2}+(0-(-4))^{2}}=\sqrt{52}=2\sqrt{13}\\\\\therefore \overline{AB} \cong \overline{CD}$

Since ABCD has two pairs of opposite congruent sides, it is a parallelogram.

3 0

2 years ago

Standard form for -10-6x^4-7x^2+5x

qaws [65]

-6x^4-7x^2+5x-10. hope this helps ! :)

7 0

2 years ago