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3 years ago

On a cold day in Moscow, the temperature was 5∘C. If the temperature drops 17∘, what is the new temperature?

Mathematics

2 answers:

Ksju [112]3 years ago

5 0

Answer:

5-15 equals -12

so the temp was -12

suter [353]3 years ago

4 0

5-10=-5 another -7 =-12

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Help quick or else trouble

sasho [114]

Answer:

1/21

Step-by-step explanation:

2/7 ÷ 6 =

2/7 x 1/6 = 2/42

Simplify by dividing the numerator and denominator by 2

1/21

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

7 0

3 years ago

URGENT!!!! WILL GIVE BRANLIEST !!!! Find the height, x, of the building. A) 113.3 ft B) 91.8 ft C) 44.5 ft D) 233.7 ft

ASHA 777 [7]

Answer:

91.8 ft

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan theta = opp/ adj

tan 42 = x/102

102 tan 42 =x

91.84121252 =x

5 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=Y%3D%5Cfrac%7B-2%7D%7Bx%7D%2B4" id="TexFormula1" title="Y=\frac{-2}{x}+4" alt="Y=\frac{-2}{x}+

Studentka2010 [4]

Part A: Vertical asymptote is $x=0$

Part B: Domain is $\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)$

Part C: Horizontal asymptote is $y=4$

Part D: Range is $\left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)$

Explanation:

Part A: We need to determine the vertical asymptote

The vertical asymptote of a function can be determined by equating the denominator equal to zero.

Thus, we have,

$x=0$

Hence, the vertical asymptote is $x=0$

Part B: We need to determine the domain

The domain of the function is the set of all independent x - values for which the function is real and well defined.

Let us take the denominator and equate to zero.

Hence, we have, $x=0$

Therefore, the function is undefined at the point $x=0$

Thus, the domain of the function is $\left(-\infty \:,\:0\right)\cup \left(0,\:\infty \:\right)$

Part C: We need to determine the horizontal asymptote

The horizontal asymptote of the function can be determined by dividing the leading coefficient of the numerator by leading coefficient of the denominator.

Thus, we have, $y=4$

Hence, the horizontal asymptote of the function is $y=4$

Part D: We need to determine the range

The range of the function is the set of all dependent y -values of the function.

In other words, the range of the function can be determined by substituting the values for x.

Thus, we have,

$\left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)$

Therefore, the range of the function is $\left(-\infty \:,\:4\right)\cup \left(4,\:\infty \:\right)$

6 0

3 years ago

She paid 26.35 for something that would be 31 what percent discount did she get

vladimir2022 [97]

(Markup %) = $21,250.

4 0

3 years ago

Shapes A and B are similar.

kari74 [83]

The scale factor from A to B is 5/3 and the value of r is 33/5

<h3>The scale factor from A to B</h3>

From the figure, we have the following corresponding sides

A : B = 5 : 3

Express as fraction

B/A = 3/5

This means that, the scale factor from A to B is 5/3

<h3>The value of r</h3>

From the figure, we have the following corresponding sides

A : B = 11 : r

Express as fraction

B/A = r/11

Recall that:

B/A = 3/5

So, we have:

3/5 = r/11

Multiply by 11

r = 33/5

Hence, the value of r is 33/5