All categories

2 years ago

Plz help,.................refresh the page before u answer\

Mathematics

1 answer:

Eddi Din [679]2 years ago

5 0

So if p then q
the converse is if q then p

so
if x is odd then 2x+1 is odd
converse is
if 2x+1 is odd, then x is odd

4th option

You might be interested in

Which graph can be used to find the solution(s) to 4x=-4x?

Elden [556K]

<h2>Answer:</h2>

The graph is shown in the Figure below

<h2>Step-by-step explanation:</h2>

In this exercise, we have an equation. On the left side we have a straight line with slope $m=4$ and there is no any y-intercept. On the right side, on the other had, we also have a straight line, but the slope here is $m=-4$ . Therefore, by plotting these two straight lines, we have that the solution is the origin, that is, the point $(0, 0)$ .

6 0

2 years ago

Read 2 more answers

Please answer this correctly

Elan Coil [88]

Answer:

t=13

Step-by-step explanation:

the ratio of the shorter sides is t/4 and the ratio of the longer sides is 26/8 so t=26×4/8=104/8=13

8 0

2 years ago

Read 2 more answers

A baseball team ordered 12 jerseys. Each jersey costs the same amount. The shipping cost for the jerseys was $3 and the total co

alexandr402 [8]

Answer:

The answer is D if you forgot the 3 between the = sign and the + sign

Step-by-step explanation:

I just did 12 x p + 3 = 99

You have to do this because 12 times the amount of jerseys ordered (p) plus the 3 dollars of extra shipping cost. That equals 99.

3 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$