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2 years ago

Vita has 8 times more candy than Anna, but Uldis has 10 candies more than Anna.

Mathematics

2 answers:

Gekata [30.6K]2 years ago

5 0

Your mom dad grandma grandpa aunt uncle step dad or step mom or foster parents

VARVARA [1.3K]2 years ago

3 0

Answer: 19

Step-by-step explanation:

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Paco uses a spinner to select a number from 1 through 5, each with equal probability. Manu uses a different spinner to select a

lions [1.4K]

There are 5 x 10 = 50 different outcomes.

Of them only 3x10, 4x10, 4x9, 4x8, 5x10, 5x9, 5x8, 5x7 and 5x6 are greater or equal than 30. Those are 9 possibilities.

Then 50 - 9 = 41 are the possibilities that the product of the two numbers is less than 30.

The probalility, then, is 41/50 = 0.82

7 0

3 years ago

Read 2 more answers

Only need to answer question 1 but if you can pls answer both

algol [13]

Use the app photo map

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3 years ago

What is the value of x to the nearest tenth?

jenyasd209 [6]

The segment of length x bisects the chord of length 19.2.
The diameter is 24, so the radius is 12.
You have a right triangle with hypotenuse 12, and one leg 19.2/2 = 9.6

9.6^2 + x^2 = 12^2

92.16 + x^2 = 144

x^2 = 51.84

x = 7.2

3 0

2 years ago

Write the expression as either the sine, cosine, or tangent of a single angle. tan5x - tan2y / 1 + tan5x tan2y

Kryger [21]

$\bf \textit{Sum and Difference Identities}
\\\\
tan(\alpha + \beta) = \cfrac{tan(\alpha)+ tan(\beta)}{1- tan(\alpha)tan(\beta)}
\qquad
tan(\alpha - \beta) = \cfrac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)}\\\\
-------------------------------\\\\
\cfrac{tan(5x)-tan(2y)}{1+tan(5x)tan(2y)}\implies tan(5x-2y)$

6 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,

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

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago