All categories

3 years ago

Can someone plz help me with this?

Mathematics

1 answer:

Yuki888 [10]3 years ago

3 0

Option B: 3362

Option D: 6332

Option E: 4533

Solution:

To determine in which numbers are the value of the 3 on the left is 10 times the value of the 3 on the right:

Option A: 3623

The value of 3 in the left = 3000

The value of 3 in the right = 3

3 × 10 = 30

It is not correct.

Option B: 3362

The value of 3 in the left = 3000

The value of 3 in the right = 300

300 × 10 = 3000

It is correct.

Option C: 6323

The value of 3 in the left = 300

The value of 3 in the right = 3

3 × 10 = 30

It is not correct.

Option D: 6332

The value of 3 in the left = 300

The value of 3 in the right = 30

30 × 10 = 300

It is correct.

Option E: 4533

The value of 3 in the left = 30

The value of 3 in the right = 3

3 × 10 = 30

It is correct.

Option F: 3939

The value of 3 in the left = 3000

The value of 3 in the right = 30

30 × 10 = 300

It is not correct.

Therefore, the value of the 3 on the left is 10 times the value of 3 in the right numbers are

Option B: 3362

Option D: 6332

Option E: 4533

You might be interested in

A line passes through the points (3,5) and (2,0) what is the slope of the line?

Veronika [31]

Answer:

Step-by-step explanation:

slope = (difference in y)/(difference in x)

subtract the y-coordinates: 5 - 0 = 5

subtract the x-coordinates in the same order: 3 - 2 = 1

Divide the difference in y by the difference in x.

slope = 5/1 = 5

4 0

3 years ago

Triangle EFG has the following vertices E(1,2), F(4,2) and G(1,6). What are the new coordinates of triangle E’F’G’ after it is r

Fynjy0 [20]

Answer:

i dont know but

Step-by-step explanation:

DO ANYTHING FO CLOUTTTTTTTTTT

7 0

2 years ago

Evaluate −12(−2). 14 −14 −24 24

bixtya [17]

The answer would be positive 24. The reason being is because of the fact that two negatives when multiplied cancel the negative. Also 12 * 2 is 24 so yeah

5 0

3 years ago

Read 2 more answers

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft

lakkis [162]

$\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}$

$\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}$

$\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}$

8 0

3 years ago

I have the first part but I don't know how to do the second part h e l p

Debora [2.8K]

$\bf \begin{array}{cll} \stackrel{ordinal}{position}&\stackrel{term}{value}\\ \cline{1-2} 1&12\\ 2&12+d\\ 3&12+d+d\\ 4&12+d+d+d\\ &12+3d \end{array}\qquad \implies ~\hfill \begin{array}{llll} \stackrel{\textit{4th term}}{12+3d}~~=~~\stackrel{\textit{4th value}}{3}\\\\ 3d=-9\implies d = \cfrac{-9}{3}\\\\ \stackrel{\textit{common difference}}{d = -3} \end{array}$

well, we know the common difference is -3, to go from the 4th term to the 8th term, we need to add "d" 4 times or namely 3+4(-3), likewise to go from the 13th term to the 19th term we have to add "d" 6 times, or namely -24 + 6(-3).

$\bf \stackrel{\textit{8th term}}{3+4(-3)}\implies 3-12\implies -9 \\\\\\ \stackrel{\textit{19th term}}{-24 + 6(-3)}\implies -24-18\implies -42$

3 0

3 years ago