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3 years ago

This isn’t a math question but I need help!!!

1 answer:

6 0

Well he probably likes you and if you like him tell him, but make sure he isn’t playing you and you are ok with dating first. If you don’t like him and he likes you let him down nicely cause you don’t wanna be rude. All the signs are there, so do what ever pleases you.

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There are four paths from the cabin to the lake. How many ways can you walk from the cabin to the lake and back?

IceJOKER [234]

.............................................your answer is 16

3 0

4 years ago

Read 2 more answers

4. Find the coordinates in the table after a translation 3 units right and 4 units down followed by a reflection over the x-axis

never [62]

Answer:

translation 3 units right and 4 units down (7, -6)

reflect over x-axis (7, 6)

Step-by-step explanation:

The point is (4, -2), you translate the domain (4) 3 to the right which makes the domain 7, then translate 4 units down to get the range. The range is now -6 because -2 -4= - 6. To reflect over the x- axis, count upward from the point until you get to the 7, 0, and then count 6 more upward to get (7, 6) the point after it was reflected over the x-axis

8 0

3 years ago

7. The hexagon GIKMPR is regular. The dashed line segments form 30 degree angles. what is the image of oh after a rotation of 18

velikii [3]

Answer:

The image of OH is ON.

Step-by-step explanation:

Te figure GIKMPR is a regular hexagon. The number of vertices of a regular hexagon is 6. The central angle between any two consecutive vertices is 60 degree.

The dashed line segments form 30 degree angles.

If we rotate the hexagon 180 degree about O, then the each point shifts to 6th place from its original place.

Since we rotate the hexagon 180 degree about O, so the image and preimage lies on a straight line. Because a straight line make angle of 180 degree.

The line OH and ON lies on a straight line therefore the image of OH is ON.

3 0

4 years ago

Derivative of these questions<br>

shtirl [24]

Answer: $\bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}$

<u>Step-by-step explanation:</u>

$y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}$

$\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}$

$\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)$

3 0

3 years ago

Draw a graph of y=4x-1 on the grid

I am Lyosha [343]

Step-by-step explanation:

......................

5 0

3 years ago

Read 2 more answers