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3 years ago

What is √7 rounded to the nearest hundredth?

Mathematics

1 answer:

Rama09 [41]3 years ago

4 0

Answer:

2.64

Step-by-step explanation:

The sqhate root of 7 is 2.64575131106. Usually this would be 2.64 when rounded to nearest hundredth.

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A conclusion which is arrived at by Inductive reasoning is called a _____?

Ann [662]

Answer: A conclusion you reach using inductive reasoning is called conjecture.

Step-by-step explanation: Examining several specific situations to arrive at a conjecture is called inductive reasoning.

5 0

3 years ago

PLEASE HELP, THIS IS REALLY IMPORTANT. PLEASE ANSWER ASAP.

Assoli18 [71]

Yes the diagonals of a parallelogram have the same midpoint since they ... of the intersection of the diagonals of parallelogram AB CD given the vertex points ... If a parallelogram is a rhombus then its diagonals are? , statement 2 is the answer

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3 years ago

Find the value of f(g(-2))

PSYCHO15rus [73]

Answer:

The correct ans is 3.

g(-2) = 2(-2) + 5

=-4 + 5 = 1

f(g(-2)) = 4 - (1)^2

= 4 - 1

Step-by-step explanation:

4 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago

Find the number between 0.0004 and 0.0005

Margarita [4]

Answer:

0.00045

Step-by-step explanation:

This decimal is in the middle of these two numbers.

5 0

2 years ago