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3 years ago

MORE EASY 7TH GRADE MATH X/2=95 THANK YOU

Mathematics

2 answers:

lara31 [8.8K]3 years ago

6 0

Answer: 190

Step-by-step explanation:

You have to work backwards so you can form an equation like this and solve

95 x 2 = x

95 x 2 = 190]

x = 190

You can check by doing 190 divided by 2 which is 95.

SIZIF [17.4K]3 years ago

4 0

Answer:

X/2=95

X= 95*2

X=190

Step-by-step explanation:

can i have brainliest please

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3 years ago

Reasoning Graph the set of ordered pairs (0, 2). (1.4), (2,6). (3.8). Determine whether the relationship is a linear function Ex

polet [3.4K]

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No, it does not. See below.

Step-by-step explanation:

Lets find out the linear equation that passes trough (-2, 0) and (0, -6).

We know every linear equation has the form: y = mx + b

Where m is the slope on the curve and b the independent term.

We know that, given 2 points (x1,y1) and (x2,y2) we can find the slope m as:

m = (y2-y1)/(x2-x1)

In our case lets replace (x1,y1) and (x2,y2) by (-2, 0) and (0, -6) (notice it could be done in the inverse sense where (-2, 0) is (x2,y2) and (0, -6) is (x1,y1) ). So, our slope is:

m = [-6 - 0] / [0 - (-2)]

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So, we have a downward linear function with slope -3, this is:

y = -3x + b

Now, for finding b just replace any of the 2 points given in the equation. Lets replace (-2, 0):

0 = -3(-2) + b

0 = 6 + b

Subtracting 6 in both sides:

-6 = b

So, our independent term is -6 and the function is:

y = -3x - 6

Now lets see if this linear eqution passes trough (2,6). If it does, we can replace the values on the equation. Replacing x by 2:

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So, in our equation, we x is 2 y is -12, and not 6 as in the point (2,6). So, our equation does not passes trough (2,6)

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3 years ago

Look at the equation below. Identify which equation has no solution

IgorC [24]

Answer:

Step-by-step explanation:

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2 years ago

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago