All categories

4 years ago

HELPPPPP PLEASE!!

1 answer:

7 0

The answer is D.
To solve it, you have to add the fractions.
1) Since it is in mixed numbers, you have to turn the fractions to improper.
   2 1/2 + 1 1/4 = 5/2 + 5/4

2) Next find the lowest common multiples of the denominators( 2 and 4)
2: 2, <u>4
</u> 4: <u>4
</u>
3) Solve
<u> 5</u> x 2
2 x 2 = 10/4  10/4 + 5/4 = 15/4 <--make sure you only add the nominators after you find the common multiples.

4) Turn into a proper fraction
15/4 = 3 3/4

There are 3 3/4 gallons of paint all together. Therefore, answer D is correct.

You might be interested in

Expand the following: 6(x – 4y)

Lemur [1.5K]

Answer:

6(x-4y) = 6x - 24y

Step-by-step explanation:

Here, we want to expand the given expression

We can do this using the distributive approach

We have this as;

(6 * x) - (6 * 4y)

= 6x - 24y

3 0

3 years ago

7. Mrs. Golden pulls names from the jar to determine who will be the class leader. There are 13

Artist 52 [7]

Answer:

9:22

Step-by-step explanation:

9 is the boys 13 is the girls

22 is the total

3 0

3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

6 0

3 years ago

Give an example of two whole number factors with a product of 9

Monica [59]

3 and 3 or 1 and 9 not really sure of anything else

7 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers