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3 years ago

I need help how to do this problem

Mathematics

1 answer:

butalik [34]3 years ago

5 0

This is a simple ratio question, but first you need to find the number of non red marbles uses... call it a

Jar Tank
54 a
----- ----
1 6750

so using your multiplication skills the number of non red marbles is

Non red = 54 x 9750

thene the total number of marbles in the tank is Red + Non Red

hope it helps

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One afternoon in January, the temperature outside was -4*F. By midnight, the temperature had dropped by 10*F. what was the tempe

Leokris [45]

Answer:

i think the answer would be -14°F

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2 years ago

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Which of these expressions is equivalent to log(25 • 18)?

riadik2000 [5.3K]

Answer: B. log(25) + log(18)

Step-by-step explanation:

Use the rule for the sum of logs: the log of a product is the sum of the logs, which means log(x) + log(y) = log(xy) if they have the same base.

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3 years ago

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serious [3.7K]

Set $f(x)=2x^3-3x^2+2$ . Find the tangent line $\ell_1(x)$ to $f(x)$ at the point when $x=x_1$ :

$f'(x)=6x^2-6x\implies f'(x_1)=12$ (slope of $\ell_1$ )

$\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9$

Set $x_2=-\dfrac9{12}$ , the root of $\ell_1(x)$ . The tangent line $\ell_2(x)$ to $f(x)$ at $x=x_2$ has slope and thus equation

$f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}$

which has its root at $x_3=-\dfrac{151}{224}\approx-0.6741$ .

(The actual value of this root is about -0.6777)

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3 years ago

Write the given expression in terms of x and y only.<br> sin(sin−1(x) + cos−1(y))

yan [13]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2308127

_______________

Write the expression below in terms of x and y only:

(I'm going to call it "E")

$\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}$

Let

$\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}$

so the expression becomes

$\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}$

•   Finding $\mathsf{sin\,\alpha:}$

$\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}$

•   Finding $\mathsf{cos\,\alpha:}$

$\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}$

because $\mathsf{cos\,\alpha}$ is positive for $\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}$

•   Finding $\mathsf{cos\,\beta:}$

$\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}$

•   Finding $\mathsf{sin\,\beta:}$

$\mathsf{cos^2\,\alpha=y^2}\\\\
 \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ 
\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}$

because $\mathsf{sin\,\beta}$ is positive for $\mathsf{\beta\in [0,\,\pi].}$

Finally, you get

$\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}$

I hope this helps. =)

Tags:   <em>inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry</em>

6 0

3 years ago

I NEED HELP PLEASE WILL MARK BRAINLIEST IF I CANNNNN

choli [55]

Answer:153.9

Step-by-step explanation:

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3 years ago

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