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2 years ago

3 3/8 - 1 3/4 how do I subtract this

Mathematics

2 answers:

Dmitry_Shevchenko [17]2 years ago

8 0

You take 13-8 then you take that answer minus 4 then you take that answer minus botg of the 3' s

alexandr1967 [171]2 years ago

5 0

You make the denominators into common factors (the same numbers( and subtract them like that

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Why is the percent increase of 45 to 75 not equal to the percentage crease of 75 to 45 check all that apply.

nevsk [136]

Answer: B and F

Step-by-step explanation:

Since, In the first case original amount = 45

So, the increasing percentage when the original amount increases to 75,

= $\frac{75-45}{45} \times 100 = \frac{30}{45} \times 100= \frac{200}{3}$ %

In second case original amount = 75

So the decreasing percentage when the original amount decreases to 45,

= $\frac{75-45}{75} \times 100 = \frac{30}{75} \times 100= \frac{40}{1}$ %

Thus, We found that the ratio of the percent increase is not the same as the percent decrease ( because, $\frac{200}{3}\neq \frac{40}{1}$

Also, The original amount of the present increase is different from the original amount for the persons decrease. ( Because, $75\neq 45$

Therefore, Option B and F is correct only.

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2 years ago

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Ruby has 4 feet of rope that she wants to cut into 5 equal-sized pieces

Sonbull [250]

Answer:

Each length of the rope will be 0.8 feet long

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2 years ago

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How many kilograms are equivalent to 3,200 grams???

zzz [600]

The answer is 3.2
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2 years ago

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Please help me with this

victus00 [196]

Yes, because there are two pairs of congruent corresponding angles

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2 years ago

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assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

2 years ago