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3 years ago

1.the science club went on a two day field trip the first day the members

1 answer:

8 0

1. y=60n+15n y=95n+12n
2. y=15x+2
3. Equations that require multiple steps solve real world problems. This involves translating words into algebraic equations and solving them.

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Plllllllease helpppppp

Anna007 [38]

Answer:

Step-by-step explanation:

b/c we know that these triangles both have equal sides... that is given that ab and be are the same length. and that be and cd are parallel , we know that they both are isosceles triangles and that the base angles are the same. The side on ad and ae have equal angles.

so we can make the equation

2a +56 = 180 (b/c we know that around a triangle it's 180°

2 a = 124

a = 62

so ∠ BAE = 62°

3 0

3 years ago

300-7 [4 (3+5)]+3 to the 3rd power

Allisa [31]

The expression to solve is the
300-7 [4 (3+5)] + 3 to the 3rd power
3 to the third power means 3³, so
300-7 [4(3+5)]+3³
= 300 - 7 [4(8)] + 27
= 300 - 7[32] + 27
= 300 - 224 + 27
= 76 + 27
= 103
so, by solving this we get 103

4 0

3 years ago

Which of the following square root of -80

Marta_Voda [28]

You can factor -80 as

$-80 = (-1)\cdot 16 \cdot 5$

So, we have

$\sqrt{-80} = \sqrt{(-1)\cdot 16 \cdot 5}$

The square root of a product is the product of the square roots:

$\sqrt{-80} = \sqrt{(-1)}\sqrt{16}\sqrt{5}$

Since $i^2=-1$ and $4^2=16$ , we have

$\sqrt{-80} = 4i\sqrt{5}$

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

2 years ago

What is the slope of the line on the graph below?

agasfer [191]

Answer:

Your answer will be one-third. I hope this will help you.

Step-by-step explanation:

6 0

3 years ago

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