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2 years ago

I need help ASAP, would really appreciate it, marking first as brainliest.

Mathematics

2 answers:

NNADVOKAT [17]2 years ago

6 0

Answer:

A and B

Step-by-step explanation:

The lines are both in slope intercept form:

y=Mx+b

Where m is the slope and b is the y intercept.

We have 2 lines:

y=1/3x+3

y=1/3x-2

1/3 is the coefficient of x, which means the slope is 1/3. Both of these lines have the same slope of 1/3. Since they both have the same slope, they are also parallel lines.

Therefore, A: the same slopes and B: parallel lines is correct.

In the photo, the red line is 1/3x+3 and the blue like is 1/3x-2

swat322 years ago

3 0

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nasty-shy [4]

Answer:

2239grams

Step-by-step explanation:

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3 years ago

Please Help. If ∠P is given as well as the values of p, q, and r, then explain whether the Law of Sines or the Law of Cosines sh

sertanlavr [38]

Answer:

Both Law of Sines and Cosines can be used to determine the angle Q.

Step-by-step explanation:.

Since from Law of Sines with one angle and three sides we can find other angles using the ratio obtained with the given angle and side length opposite side if angle P is not given we couldn't use this.

Law of Cosines can be used to find any angle of triangle with all three side lengths given and angle P is also not required to find angle Q.

3 0

3 years ago

Read 2 more answers

4100 grams because 1 kilogram would = 100 grams

LenaWriter [7]

4. 1 kg because you will have to divide 4100 by 1000. Ps. 1 kg is equal to 1000 grams.

6 0

3 years ago

At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as

MAXImum [283]

<u>Answer-</u>

At $x= \frac{1}{2304e^4-16e^2}$ the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

$K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}$

Here,

$y=4e^{x}$

Then,

${y}' = 4e^{x} \ and \ {y}''=4e^{x}$

Putting the values,

$K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}$

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

${k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}$