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3 years ago

At 6 am the temperature outside was -7 degrees. By 2 pm it had warmed up to 18 by how many degress did the temperature change

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

5 0

Answer:

25 degrees -7 + 25 = 18

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Help me with this! I'll give Brainliest!!!!!!!!!!!!!!!

kirill115 [55]

Liana is correct bc henry included negative numbers

6 0

3 years ago

Read 2 more answers

A negative number divided by a positive number is:

Verdich [7]

Answer:

When you divide a negative number by a positive number then the quotient is negative.

Step-by-step explanation:

hope this helped

7 0

2 years ago

I need help

Nadusha1986 [10]

Answer:

f(g(x)) = 4x² - 12x + 9

Step-by-step explanation:

Substitute x = g(x) into f(x) , that is

f(g(x))

= f(2x - 3)

= (2x - 3)² ← expand using FOIL

= 4x² - 12x + 9

4 0

2 years ago

3 1 - 2.6 ________ 0.05

olya-2409 [2.1K]

Work:

$\frac{31 - 2.6}{0.05}$

$\frac{28.4}{0.05}$

Answer:

$567. \frac{}{9}$

6 0

3 years ago

Can someone please help me with my Question #29 of The Quadratic Relations for me please?

lara [203]

Answer:

Calculate the first differences between the y-values:

$\sf 3 \underset{+1}{\longrightarrow} 4 \underset{+3}{\longrightarrow} 7 \underset{+5}{\longrightarrow} 12 \underset{+7}{\longrightarrow} 19$

As the first differences are not the same, we need to calculate the second differences:

$\sf 1 \underset{+2}{\longrightarrow} 3 \underset{+2}{\longrightarrow} 5 \underset{+2}{\longrightarrow} 7$

As the second differences are the same, the relationship between the variable is quadratic and will contain an $x^2$ term.

--------------------------------------------------------------------------------------------------

To determine the quadratic equation

The coefficient of $x^2$ is always half of the second difference.

As the second difference is 2, and half of 2 is 1, the coefficient of $x^2$ is 1.

The standard form of a quadratic equation is: $y=ax^2+bx+c$

(where a, b and c are constants to be found).

We have already determined that the coefficient of $x^2$ is 1.

Therefore, a = 1

From the given table, when $x=0$ , $y=12$ .

$\implies a(0)^2+b(0)+c=12$

$\implies c=12$

Finally, to find b, substitute the found values of a and c into the equation, then substitute one of the ordered pairs from the given table:

$\begin{aligned}\implies x^2+bx+12 & = y\\ \textsf{at }(1,19) \implies (1)^2+b(1)+12 & = 19\\ 1+b+12 & = 19\\b+13 & =19\\b&=6\end{aligned}$

Therefore, the quadratic equation for the given ordered pairs is:

$y=x^2+6x+12$

7 0

2 years ago