Help Please ..................

Mathematics

2 answers:

Paul [167]2 years ago

7 0

Plug in -11 for x
8(-11) + 2.
-86

Studentka2010 [4]2 years ago

5 0

Answer:

-86

Step-by-step explanation:

Substitute -11 for x in the equation:

8(-11)+2

=-88+2

=-86

You might be interested in

Which one of the following is considered a quantitative variable?

tatyana61 [14]

Okay so here's the definition of quantitative just in case you ever need it :) Categorical. Categorical variables take on values that are names or labels. The color of a ball (e.g., red, green, blue) or the breed of a dog (e.g., collie, shepherd, terrier) would be examples of categorical variables. And the answer would be A the color of your car because it talks about in the definition of the color of a ball so A would be the correct answer! Please mark brainliest :)

7 0

2 years ago

Helpppppppppppp ppppp:/

jeka94

Answer:

Step-by-step explanation:

I hope this helps

3 0

3 years ago

I have one qt of solution diluted out 1:4 How much do I need to add to end up with a final product that’s a 1:10 solution?

LenaWriter [7]

Answer:

1.1352 L

Step-by-step explanation:

L = qt 1.0567

qt = 0.946 L

1:4 = 5

0.946/5 = 0.1892 L one part

0.1892 x 11 = 2.0812 11 parts as 1+10 =11

Answer therefore is subtracting 5 from 11 parts to get the final 6 parts added = 2.0812 - 0.946

= 1.1352 L

7 0

2 years ago

Every $4,000 spent on cable TV will reach 160 fans while every $2,750 spent on radio ad will reach 125 fans. A stadium has estab

laila [671]

Answer:

Neither of the two advertising options meet the goal.

Step-by-step explanation:

Every $4,000 spent on cable TV will reach 160 fans.

Therefore, the goal of reaching 200000 fans will be fulfilled by $\frac{4000 \times 200000}{160} = 5000000$ dollars i.e. $5 million which is greater than the budget of $3 million.

Now, every $2,750 spent on radio ads will reach 125 fans.

Therefore, the goal of reaching 200000 fans will be fulfilled by $\frac{2750 \times 200000}{125} = 4500000$ dollars i.e. $4.5 million which is again greater than the budget of $3 million.