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2 years ago

Decimal rounded to the nearest hundredth is 6.32. This decimal is greater than 6.32. What could this decimal be?

Mathematics

1 answer:

drek231 [11]2 years ago

6 0

6.320 add a place holder, hope this helps

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Two polynomials are given below.

Kazeer [188]

P+q= 5x^2+2
q-p= -(x-2) (x+2)

3 0

2 years ago

Which equation can be solved by using this expression?

Marat540 [252]

Answer:

The correct answer option is B. 2 = 3x + 10x^2

Step-by-step explanation:

We are to determine whether which of the given equations in the answer options can be solved using the following expression:

$x=\frac{-3 \pm\sqrt{(3)^2+4(10)(2)} }{2(10)}$

Here, $a = 10, b = 3$ and $c=-2$ .

These requirements are fulfilled by the equation 4 which is:

$12=3x+10x^2$

Rearranging it to get:

$10x^2+3x-2=0$

Substituting these values of $a,b,c$ in the quadratic formula:

$x= \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

$x= \frac{-3 \pm\sqrt{(3)^2-4(10)(-2)} }{y}$

3 0

2 years ago

At noon, the minute and hour hands overlap. In how many hours will they overlap again?

nadya68 [22]

24 hours

Because 24 hours equals 1 day or one rotation of the clock

7 0

2 years ago

Cubed root x cubed root x2

Yuki888 [10]

Answer:

Final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

Step-by-step explanation:

Given problem is $\sqrt[3]{x}\cdot\sqrt[3]{x^2}$ .

Now we need to simplify this problem.

$\sqrt[3]{x}\cdot\sqrt[3]{x^2}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}$

Apply formula

$\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}$

so we get:

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$

Hence final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

7 0

2 years ago

(I NEED THIS ANSWERED SUPER SUPER FAST PLEASE IT'S IMPORTANT!!!!!!) 6 7/8c =__ fl oz

Olegator [25]

1 cup = 8 fluid ounces.
6.875 cups = ? fluid ounces.
8*6.875=55.

Your answer would be 55 fluid ounces.

8 0

2 years ago