All categories

3 years ago

Write a real word problem in which you would divide decimals.

Mathematics

2 answers:

Viefleur [7K]3 years ago

7 0

8.54 / 0.7 =12.2 is one i guess

irinina [24]3 years ago

5 0

Erica visits her grandmother once every 2.5 years. If Erica is 30 years old, how many times has she visited her grandmother.

We know that Erica visits her grandmother once every 2.5 years and that Erica is 30 years old. We want to find out how many times she has visited her grandmother.

So we need to divide Erica's age (30 years) by how frequently she visits (one every 2.5 years.) Notice that we are dividing by a decimal so in this situation, we need to change 2.5 to a whole number by multiplying both 2.5 and 30 by 10 so that we have 300 ÷ 25.

From here, we find that 25 divides into 100 exactly 12 times so we know that Erica has visited her grandmother 12 times.

I attached my work in the image provided.

You might be interested in

The Antonov An-225 is an extremely large plane. Its length is 276 ft. – about 2½ football fields! A model of the plane has scale

Mademuasel [1]

Answer:

You can download the answer here

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

When you are factoring a trinomial with a leading coefficient other than 1, which would be the best thing to do first?

Alchen [17]

Answer:

Step-by-step explanation:

c. Look for a common factor in each term.

4 0

3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$