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3 years ago

7

A particular plant root grows 1.5 inches per month. How many centimeters is the plant root growing per month? (1 inch = 2.54 cen

timeters).

2 answers:

Lady bird [3.3K]3 years ago

4 0

You convert inches into centimeters
2.54*1.5=3.81

The answer to this question is 3.81

zloy xaker [14]3 years ago

4 0

Answer:

3.81

Step-by-step explanation:

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Does 10p/2q = 5p/q<br> I think it is not equal but it doesn't seem right.

Rufina [12.5K]

Answer: $\frac{10p}{2q}=\frac{5p}{q}$ (TRUE)

Step-by-step explanation:

For this exercise it is important to remember that, by definition, fractions have the following form:

$\frac{a}{b}$

Where "a" is the numerator and "b" is the denominator.

The numerator and the denominator are Integers, but the denominator cannot be zero ( $b\neq 0$ )

In this case you have the following equation provided in the exercise:

$\frac{10p}{2q}=\frac{5p}{q}$

To find out if the left side of the equation is equal to the right side, it is necessary to simplify the fraction on the left.

Notice that the numerator and the denominator of the fraction on the left can be both divided by 2. Then, you can simplify it.

So, you get;

$\frac{5p}{q}=\frac{5p}{q}\ (TRUE)$

6 0

4 years ago

Is this graph a function or not a function

Scrat [10]

Answer:

YEs it is a function

Step-by-step explanation:

5 0

4 years ago

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0

3 years ago

HELP please! ‍♂️<br><br> Last one is 0.5 < x < 1.5<br><br> And none

Ket [755]

1.5 is lesser than 0.5. Oh, wait nevermind I meant to say -1.5 is lesser than 0.5

3 0

3 years ago

An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situati

Nostrana [21]

Answer:

Question

An expression can often be used to model a real-life situation. For each of the following expressions, write a real-life situation using money as the context. Then, evaluate the expression and interpret the result in terms of the situation. Here's an example.

The following situation is modeled by the expression -24.50 + 100.00: Jonathan overdrew his checking account by $24.50. He then transferred $100.00 from his savings account to his checking account. His new checking account total is $75.50.

Make sure to use complete sentences in your answers.

1) $8.15 + $11.99

2) $20 - $18.04

3) 7($1.45)

4) 42.08 / 4

It's too short. Write at least 20 characters to explain it well.

8 0

3 years ago