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docker41 [41]
3 years ago
13

Which problems will have two decimal places in the product?mark all that apply.

Mathematics
2 answers:
Vikki [24]3 years ago
8 0

Answer:

A) 5 x 0.89 and C) 5.31 x 10^0

Step-by-step explanation:

If you want two decimal places in the product, you have to multiply a number in the hundredths place and another number for a product in the hundredths. But the thing is, you have to make sure it doesn't end up in the tenth place such as 0.02 x 5 is 0.1.

So 5 x 0.89 is 4.45 and 5.31 x 10^0 is 5.31

Marat540 [252]3 years ago
7 0

Answer : The correct options are:

(A) 5 × 0.89

(C) 5.31 × 10⁰

(E) 3.2 × 4.3

Step-by-step explanation :

If we want two decimal places in the product then we have to multiply a number in the hundredths place and another number for a product in the hundredths.

As we are given that the following multiplication expressions. Now we have to determine the which expression have two decimal places in the product.

(A) 5 × 0.89

The multiplication of two number will be, 4.45

(B) 7.4 × 10

The multiplication of two number will be, 74.0

(C) 5.31 × 10⁰

The multiplication of two number will be, 5.31

(D) 6.1 × 3

The multiplication of two number will be, 18.3

(E) 3.2 × 4.3

The multiplication of two number will be, 13.76

From this we conclude that the expression that have two decimal places in the product are,

(A) 5 × 0.89

(C) 5.31 × 10⁰

(E) 3.2 × 4.3

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2762144

_______________


Let  \mathsf{\theta=cos^{-1}\!\left(\dfrac{4}{5}\right).}


\mathsf{0\le \theta\le\pi,}  because that is the range of the inverse cosine funcition.


Also,

\mathsf{cos\,\theta=cos\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{5}}\\\\\\ \mathsf{5\,cos\,\theta=4}


Square both sides and apply the fundamental trigonometric identity:

\mathsf{(5\,cos\,\theta)^2=4^2}\\\\
\mathsf{5^2\,cos^2\,\theta=4^2}\\\\
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\mathsf{25\cdot (1-sin^2\,\theta)=16}

\mathsf{25-25\,sin^2\,\theta=16}\\\\
\mathsf{25-16=25\,sin^2\,\theta}\\\\
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\mathsf{sin^2\,\theta=\dfrac{9}{25}}


\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{9}{25}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{3^2}{5^2}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{3}{5}}


But \mathsf{0\le \theta\le\pi,} which means \theta lies either in the 1st or the 2nd quadrant. So \mathsf{sin\,\theta} is a positive number:

\mathsf{sin\,\theta=\dfrac{3}{5}}\\\\\\
\therefore~~\mathsf{sin\!\left[cos^{-1}\!\left(\dfrac{4}{5}\right)\right]=\dfrac{3}{5}\qquad\quad\checkmark}


I hope this helps. =)


Tags:  <em>inverse trigonometric function cosine sine cos sin trig trigonometry</em>

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