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2 years ago

The measures 4, 5, and 7 create a right triangle. True False

Mathematics

1 answer:

Korvikt [17]2 years ago

5 0

<em>False.</em>

Hope This Helps You!

Good Luck Studying ^-^

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A type of cell triples every hour. If there are 5 cells initially, which function can be used to

laila [671]

Answer:

T(t) = 5(3)(t)

Step-by-step explanation:

There are 5 cells initially.

And it triple every hour.

T0 =5

T(t) = T0(3t)

Each cell has it's multipling factor because it's Will be multipling by 3 every hour

T(t) = 5(3)(t)

4 0

3 years ago

An election ballot asks voters to se4five city commissioners from a group of fifteen candidates. In how many wsys can this be do

svp [43]

That's "15 choose 5"

$\displaystyle{15 \choose 5} = \dfrac{15! }{5! 10!} = \dfrac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = (15/(5(3))(14/2)(13)(12/4)(11)\\\\= 7(13)(3)(11)=231(13)=3003$

Answer: 3003 ways

8 0

3 years ago

Please help if you can.

Ronch [10]

Plug -3 into the equation: w(-3)=14-6(-3)
simplify: w(-3)=14+18
solve: w(-3)=32

3 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

What is the meaning of the unknown factor and quotient

konstantin123 [22]

Unknown factor is multiplication and then quotient is division

5 0

3 years ago