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3 years ago

Determine the given equation is a linear equation write yes or no. If yes, write the equation in standard form.

Mathematics

1 answer:

kykrilka [37]3 years ago

6 0

Answer:

The equation is linear. Standard form:

7n - 6m - 4 =0

Step-by-step explanation:

Linear Equations

The linear equations are identified because all terms form a polynomial of degree 1 or zero (no variable).

The standard form of a linear equation is:

ax+by+cz+d=0

where x, y, z are the variables, and a,b,c are constants.

The equation:

7n - 8m = 4 - 2m

Is linear, both variables have an exponent of 1 and there is an independent term. Moving all terms to the left side:

7n - 8m - 4 + 2m =0

Simplifying:

7n - 6m - 4 =0

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Point x is located at (-2,-6) and point z is located at (0,5). Find the y value for the point y that is located 1/5 the disteanc

Sedaia [141]

Answer:

-3 1/2

Step-by-step explanation:

We have to find the distance between x and z along the y-axis.

This will be: 5 - -6 = 11 units

1/5 of 11 = 1/5 × 11

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4 years ago

Math question #1 please show steps

-Dominant- [34]

Answer:

Step-by-step explanation:

An approximation of an integral is given by:

$\displaystyle \int_a^bf(x)\, dx\approx \sum_{k=1}^nf(x_k)\Delta x\text{ where } \Delta x=\frac{b-a}{n}$

First, find Δx. Our a = 2 and b = 8:

$\displaystyle \Delta x=\frac{8-2}{n}=\frac{6}{n}$

The left endpoint is modeled with:

$x_k=a+\Delta x(k-1)$

And the right endpoint is modeled with:

$x_k=a+\Delta xk$

Since we are using a Left Riemann Sum, we will use the first equation.

Our function is:

$f(x)=\cos(x^2)$

Therefore:

$f(x_k)=\cos((a+\Delta x(k-1))^2)$

By substitution:

$\displaystyle f(x_k)=\cos((2+\frac{6}{n}(k-1))^2)$

Putting it all together:

$\displaystyle \int_2^8\cos(x^2)\, dx\approx \sum_{k=1}^{n}\Big(\cos((2+\frac{6}{n}(k-1))^2)\Big)\frac{6}{n}$

Thus, our answer is C.

*Note: Not sure why they placed the exponent outside the cosine. Perhaps it was a typo. But C will most likely be the correct answer regardless.

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3 years ago

Can you make a triangle with side lengths of 2 inches and 2 inches and 4 inches?

Grace [21]

Answer: No you can not make any triangles with 2 inches, 2 inches and 4 inches. hoped this helped!

Step-by-step explanation:

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3 years ago