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1 year ago

Find the height of a triangle if the area is 42 square units and the base is 7 units.

Mathematics

1 answer:

igor_vitrenko [27]1 year ago

3 0

Answer:

your answer is 12

Step-by-step explanation:

Base 7 Area 42

Using the formula

A= hb over 2

Solving for hb

hb=2A

b=2·42

7=12

hope this is not to confusing but the answer is 12.

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Answer: 2

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3 years ago

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How many unit tiles need to added to the expression x^2+4x+3 to form the perfect square trinomial?

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x^2 + 4x + 3 + 1 = (x + 2)^2

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2 years ago

A randomly generated list of numbers from 0 to 4 is being used to simulate an event with numbers 3 and 4 representing a success.

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3 years ago

Jackson used the process of completing the square to solve the equation 2x2−12x=−6.

Nikolay [14]

Answer:

$x=3-\sqrt{6},x=3+\sqrt{6}$

Step-by-step explanation:

we are given equation as

$2x^2-12x=-6$

Since, we have to solve it by using complete square

so, firstly we will complete square

and then we can solve for x

step-1:

Factor 2 from both sides

$2(x^2-6x)=-3\times 2$

step-2:

Simplify it

$x^2-6x=-3$

step-3:

Add both sides 3^2

$x^2-6x+3^2=-3+3^2$

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step-4:

Take sqrt both sides

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step-5:

Add both sides by 3

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3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

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