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3 years ago

Write the numeral one hundred fifty five and nine hundred two thousandths

Mathematics

1 answer:

7nadin3 [17]3 years ago

3 0

Answer:

0.15590200

Step-by-step explanation:

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ASAP! GIVING BRAINLIEST! Please read the question THEN answer CORRECTLY! NO guessing. Show your work OR give an explaination.

love history [14]

Answer:

$\boxed{D. 3 {x}^{2} - x - 6}$

Given:

$f(x) = 3{x}^{2} - 4 \\ \\ g(x) = x + 2$

To Find:

$(f - g)(x) = f(x) - g(x)$

Step-by-step explanation:

$= > f(x) - g(x) = (3{x}^{2} - 4) - (x + 2) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3{x}^{2} - 4 -(x) - (2)\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3{x}^{2} - 4 - x - 2\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3 {x}^{2} - x - 6$

8 0

3 years ago

Solve the equation for y.<br><br> 4x – 5y = 9

Brums [2.3K]

4 x - 5 y = 9
- 5 y = - 4 x + 9 / · ( - 1 )
5 y = 4 x - 9 / : 5
$y = \frac{4x-9}{5}$

7 0

4 years ago

Please help hahahehhehe

elena55 [62]

150/3 = 50

1 rocket takes 50 gallons of fuel.

152 * 50 = 7,600

To fuel 152 mini rockets it would take 7,600 gallons of fuel.

3 0

4 years ago

The set of numbers 1, 5, 13, and 24 contains values for m. What values of m makes the equation below true?

Likurg_2 [28]

Answer:

Step-by-step explanation:

Solve the equation first for "m"

6m + 12 = 42

6m = 30 (subtract 12)

m = 5 (Divide by 6)

5 is your answer

Hope this help! Please mark me brainliest, it helps!

7 0

2 years ago

Find the values of cosΘ and tanΘ, given that sinΘ = 8/9 and Θ is in quadrant I

Misha Larkins [42]

$sin^2 \alpha +cos^2 \alpha =1\ \ \ and\ \ \ tan \alpha = \frac{\big{sin \alpha }}{\big{cos \alpha }} \\\\sin \alpha = \frac{8}{9}\\\\ \Rightarrow\ \ \ (\frac{8}{9})^2+cos^2 \alpha =1\ \ \ \Rightarrow\ \ \ cos^2 \alpha =1- \frac{64}{81} \ \ \ \Rightarrow\ \ \ cos^2 \alpha = \frac{17}{81} \\\\ \alpha \ \in\ (0^0;90^0)\ \ \ \Rightarrow\ \ \ cos \alpha >0\ \ \ \Rightarrow\ \ \ cos \alpha = \frac{ \sqrt{17} }{9} \\\\$

$tan \alpha = \frac{8}{9}: \frac{ \sqrt{17} }{9} =\frac{8}{9}\cdot \frac{9}{\sqrt{17}}= \frac{8}{\sqrt{17}}= \frac{8\cdot \sqrt{17}}{\sqrt{17}\cdot \sqrt{17}}=\frac{8\cdot \sqrt{17}}{17}$

6 0

4 years ago