All categories

3 years ago

The sum of the ages of Martin and his son Sherwin is 47 years. Martin's age is 2 years more than 4 times the

Mathematics

1 answer:

aliina [53]3 years ago

7 0

Answer:

In the picture above it shows how to answer the questions step by step. If there's a mistake, pls inform me.

You might be interested in

AABC with vertices A(- 2, 3), B(3, - 3) , and C(- 2, - 3) is shown below.

kicyunya [14]

Answer:

ab= 7.8

ac= 6

bc=5

Step-by-step explanation:

4 0

3 years ago

I have 30 question test and need 70% to pass how many questions can be wrong

alekssr [168]

Total questions are 30.

70% of 30
=30(70/100)
=21

So, the questions can be wrong are 30-21.
=9 questions..

Thanks!

3 0

3 years ago

Giving 45 points! I need help!

Monica [59]

where is the question?

8 0

3 years ago

This is hard and I need some help

aliina [53]

Answer:

Step-by-step explanation:

Answer attached.

Hope it helps ^_^

4 0

3 years ago

Marcy has 150 to buy packages of hotdogs and hamburgers for her booth at the carnival. Grocery store she found packages of hotdo

bekas [8.4K]

Answer:

Part (A): The required equation will be: $6d+20b=150$

Part (B):The required equation will be: $d=\frac{150-20b}{6}$

Part (C):The required equation will be: $b=\frac{150-6d}{20}$

Step-by-step explanation:

Consider the provided information.

Let d represents number of packages of hot dogs and b represents the number of packages of hamburgers.

Part (a)

It is given that the cost of a package of hot dogs is $6 and cost of a package of hamburger is $20.

The required equation will be: $6d+20b=150$

Part (b) Solve the above equation for d.

Subtract 20b from both sides.

$6d=150-20b$

Divide 6 from both sides.

$d=\frac{150-20b}{6}$

The required equation will be: $d=\frac{150-20b}{6}$

Part (c) Solve the equation for b.

Subtract 6d from both sides.

$20b=150-6d$

Divide 20 from both sides, we get

$b=\frac{150-6d}{20}$

The required equation will be: $b=\frac{150-6d}{20}$

6 0

3 years ago