Ask question

Ask question

All categories

3 years ago

9

The temperature was 8 degrees below zero at 5 A.M. By 1 P.M. the temperature had risen 23 degrees. What was the temperature at 1

P.M.?

2 answers:

Serjik [45]3 years ago

8 0

Yea hea sound right up there

Bad White [126]3 years ago

7 0

Answer:

15 degrees

Step-by-step explanation:

You might be interested in

Can someone help with this

dimaraw [331]

Hey There! :)

Solve for g.
$g/20=12/48
$
Isolate the variable; Multiply both sides by 20.
$(g/20)*20=(12/48)*20$
$g=240/48$
$g=5$

Hope this helps.
-Benjamin

6 0

3 years ago

tekilochka [14]

-5-12=-17 so it is b)

8 0

4 years ago

Simplify and put in a Simplified Expression:<br> 12x + 3x2-6-6x

denpristay [2]

Answer:

3(2x+x²-2)

Step-by-step explanation:

12x+3x²-6-6x

collect like terms

6x+3x²-6

factor the expression

3(2x+x²-2)

3 0

2 years ago

Given that a x 70 = b work out the value of<br>3b/a<br><br>Your final line should say, 3b/a = ...

Leto [7]

Answer:

$\frac{3b}{a} = 210$

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given that a× 70 = b

Dividing 'a' into both sides, we get

⇒ $\frac{b}{a} = 70$

we have to find the value $\frac{3b}{a}$

$\frac{3b}{a} = 3 X \frac{b}{a}$

= 3× 70

= 210

3 0

3 years ago

Which expression is equivalent to log w (x^2 -6)^4/ 3 sqrt x^2+8?

evablogger [386]

Answer:

C $4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8)$

Step-by-step explanation:

First use the property of logarithms

$\log _ab-\log_ac=\log_a\dfrac{b}{c}.$

For the given expression you get

$\log_w\dfrac{(x^2-6)^4}{\sqrt[3]{x^2+8} }=\log_w(x^2-6)^4-\log_w\sqrt[3]{x^2+8}=\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}$

Now use property of logarithms

$\log_ab^k=k\log_ab.$

For your simplified expression, you get

$\log_w(x^2-6)^4-\log_w(x^2+8)^{\frac{1}{3}}=4\log_w(x^2-6)-\dfrac{1}{3}\log_w(x^2+8).$

3 0

3 years ago