Work out the missing number a) __ ÷ (-3)=27 b) __x(-2)=70 Hlep pls

Evaluate the expression a2+6ab for a=−3 and b=4.

inn [45]

First you plug in the variables:

(-3)(2)+(6)(-3)(4)

Multiply:
-6+(-18)(4)
-6+-72

Add and your answer comes out to be:
-78

5 0

3 years ago

PLEASE HELP!!!!! Florian ran 1.2 miles and walked 4.8 laps around the path at the park for a total distance of 3.6 miles. Which

fgiga [73]

Answer:

The correct answer would be D) 4.8x + 1.2 = 3.6; x = 0.5 mile

Step-by-step explanation:

This is because laps would be the dependent variable, so we know the number of them (4.8) would be multiplied by the variable (x). We also know that 1.2 is the constant. Now we can solve to make sure this is the right equation.

4.8x + 1.2 = 3.6

4.8x = 2.4

x = 0.5

7 0

3 years ago

Read 2 more answers

Combine the like terms to create an equivalent expression: -3r- 6+ (-1)

Norma-Jean [14]

Answer:

-3r+5 is that answer hdj

7 0

2 years ago

Read 2 more answers

Roman is born in the year 123 BC and dies at the age of 74 years. Use a negative number to express the year in which he dies

Stells [14]

Answer:

-49 or 49 BC

hope that this helps

7 0

3 years ago

DIscrete Math

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

3 0

2 years ago

Work out the missing numbera) __ ÷ (-3)=27b) __x(-2)=70Hlep pls ​

Work out the missing numbera) ÷ (-3)=27b) x(-2)=70Hlep pls