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3 years ago

Six degrees warmer than -40°F.

Mathematics

2 answers:

g100num [7]3 years ago

8 0

Answer:

-34 degrees

Step-by-step explanation:

If it were warmer, you would add so -40+6 = -34 so the answer is -34.

snow_tiger [21]3 years ago

6 0

Answer:

6 Degrees warmer than -40°F would be <em><u>-34°F</u></em>. Hope this helps you! :)

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Simplify the expression -3/5(-8+5/9x-3).

Gekata [30.6K]

Answer:

3/5(-8+5/9x-3)

3/5(-8+5x/9-3)

3/5(5x/9-3)

7 0

3 years ago

Can someone please help me and write out the answer

Elina [12.6K]

Answer:

$- 30$

Step-by-step explanation:

$\sin {}^{ - 1} ( - \frac{1}{2} ) =$

Using the identity,

$\sin( - x) = - ( \sin(x) )$

$\sin( \frac{1}{2} ) = \frac{\pi}{6}$

$\sin {}^{ - 1} ( \frac{ - 1}{2} ) = - \frac{\pi}{6}$

Convert to degrees.

$- 30$

8 0

3 years ago

HELP NOW

Brilliant_brown [7]

Answer:

$112.75

Step-by-step explanation:

5% of 118.65=5.9

118.65-5.9=112.75

6 0

3 years ago

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Donna is running a track it takes her 10 minutes to run 6 laps if she keeps running at the same speed how long will it take her

kaheart [24]

To solve this problem we can make a proportion.
6 5
-- = ---
10 x

to solve this proportion we can cross multiply and divide.
5 times 10 is 50. divided by 6 is 8.3
it will take her 8.3 minutes to run 5 laps

7 0

3 years ago

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Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

Read 2 more answers