Ask question

Ask question

All categories

3 years ago

7

Yesterday's afternoon temperature was 21°F. It plunged overnight to -3°F. How many degrees did the temperature drop? A) 3° B) 18

° C) 21° D) 24°

2 answers:

LiRa [457]3 years ago

7 0

Answer: B) 18° C

Step-by-step explanation:

21+(-3) = 21 - 3

= 18

scoray [572]3 years ago

7 0

Answer:

B) 18c

Step-by-step explanation:

21 is positive -3 is negative

celsius has

You might be interested in

Value of x in log5x = 4logx5 is

olchik [2.2K]

Log5 x=4 logx 5
Ln x/ln5=4(ln5 / ln x) logx z=ln x / ln z
least common multiple=ln5 * ln x

ln²x=4ln²5
√(ln² x)=√(4 ln²5)
ln x=2ln 5
x=e^2ln5=25

Answer: x=25.

4 0

2 years ago

If sin ⁡x=725, and 0 ∠ x ∠ pi/2, what is the tan (x - pi/4)

krok68 [10]

$Tan(x-\frac{\pi }{4}) = \frac{-17}{31}$

<u>Step-by-step explanation:</u>

Here we have , sin ⁡x=7/25( given sin x = 725 which is not possible ) , $0$ . Let's find tan (x - pi/4):

⇒ $Tanx = \frac{sinx}{cosx}$

⇒ $Tanx = \frac{sinx}{\sqrt{1-(sinx)^{2}}}$

⇒ $Tanx = \frac{\frac{7}{25}}{\sqrt{1-(\frac{7}{25})^{2}}}$

⇒ $Tanx = {\frac{7}{25}}{\sqrt{(\frac{625}{625-49})^{}}}$

⇒ $Tanx = {\frac{7}{25}}(\frac{25}{24} )$

⇒ $Tanx = {\frac{7}{24}}$

Now , $Tan(x-\frac{\pi }{4}) = \frac{Tanx - Tan(\frac{\pi }{4} )}{1+ Tanx(Tan(\frac{\pi }{4} )}$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{Tanx -1}{1+ Tanx(1)}$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{\frac{7}{24} -1}{1+\frac{7}{24} }$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{\frac{7-24}{24} }{\frac{7+24}{24} }$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{-17}{24} (\frac{24}{31} )$

⇒ $Tan(x-\frac{\pi }{4}) = \frac{-17}{31}$

7 0

3 years ago

What’s is the answer

kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

$(x^{2} )^{3} = x^6$

$(x^{3} )^5=x^{15}$

When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

$(x^{2} )(x^{3})=x^{5}$

$(x^{1} )(x^{2})=x^{3}$

$(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}$

You multiply 3 into each exponent in the numerator and the denominator

$\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}$

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

$\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}$

When you have something like this:

$\frac{x^{2}}{x^5}$

You subtract the exponents together, so:

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

Your answer is the second option

3 0

3 years ago

Pls help I would appreciate it!

anastassius [24]

Answer:

D) (2,5)

Step-by-step explanation:

The intersection of the 2 lines.

pls give brainliest.

3 0

2 years ago

One of the roots of the equation x²-5x +q=0 is 2. Find the other root and the value of the coefficient q.

yuradex [85]

Answer:

Other root: $x=3$

$q=6$

Step-by-step explanation:

If $x=2$ , then $x-2=0$ .

Therefore:

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

$2^2-5(2)+q=0$

$4-10+q=0$

$-6+q=0$

$q=6$

This means that the other root is $x=3$ since $(x-2)(x-3)=x^2-5x+6$

5 0

2 years ago