1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
zhenek [66]
3 years ago
10

The temperature went from -16°F to 7°F. What was the change in temperature? -9°F -23°F 9°F 23°F

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
5 0

Answer:

it is 23 degrees f positive       PLZ MARK AS BRAINLIEST

Step-by-step explanation:

UkoKoshka [18]3 years ago
3 0

Answer:

its 23 degrees F

Step-by-step explanation:

-16 + 23 = 7

You might be interested in
An item is regularly priced at $70. It is on sale for 70% off the regular price. What is the sale price?
Arada [10]

Answer:

$21

Step-by-step explanation:


7 0
3 years ago
$20 haircut ; 10% tip
Trava [24]

Answer:

22 dollars

Step-by-step explanation:

10% of $20= 2

20+2=$22

3 0
3 years ago
Graph Y+5=-2x and explain how you did it.
Margaret [11]
Y
=
−
2
x
+
5
y
=
-
2
x
+
5
Use the slope-intercept form to find the slope and y-intercept.
Tap for more steps...
Slope:
−
2
-
2
y-intercept:
(
0
,
5
)
(
0
,
5
)
Any line can be graphed using two points. Select two
x
x
values, and plug them into the equation to find the corresponding
y
y
values.
Tap for more steps...
x
y
0
5
5
2
0
6 0
3 years ago
What is the equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4)?
Nastasia [14]

Answer:

The equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5

Option A is correct.

Step-by-step explanation:

We need to find equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4)

We need a line parallel to x=-2 or x+2=0 it should be of form x+k=0

We need to find k, by putting value of x=-5 as given in question the point(-5,4)

-5+k=0

k=5

So, the equation of line will be found by putting k=5:

x+k=0

x+5=0

x=-5

So, the equation of the line that is parallel to the line x = -2 and passes through the point (-5, 4) is x=-5

Option A is correct.

8 0
3 years ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
Other questions:
  • Simplify 7 - 2(5 - 2x) -10x + 25 -4x - 3
    7·2 answers
  • Brainliest question please help please help me and plz show your work
    6·1 answer
  • Which inequality does this graph represent?
    5·1 answer
  • Please Help!
    9·2 answers
  • Explain how to use the combine place values strategy to find 223-119
    10·2 answers
  • What is the length of BE?<br><br> 14 units<br> 17 units<br> 27 units<br> 34 units
    9·2 answers
  • I beg anybody please
    14·1 answer
  • I need help on 4-4a please
    14·1 answer
  • A rope is 2.4 meters long . What is it's length in centimeters
    11·2 answers
  • For the function g(x) = -4x + 6, what input value results in an output value of 6?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!