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3 years ago

Estimate the product. 3 1/3 ×4 1/8 = ?

Mathematics

2 answers:

irinina [24]3 years ago

8 0

Answer:

Step-by-step explanation:

Estimation:

3 1/3 --> 3

4 1/8 --> 4

3 x 4 ~ 12

Hope that helps!

Ray Of Light [21]3 years ago

6 0

Answer:

The answer would be 13 3/4

Step-by-step explanation:

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Please help asap:) 30points

pishuonlain [190]

Answer:

q#1 Option B.2 possible solution is correct option

Q#2 option c. 1 viable solution is correct option.

Step-by-step explanation:

Q#1

y=4 $x^{2}$ +3x+45

as this is a quadratic solution

and we know that when we solve a quadratic equation then it gives two possible solutions

hence option b is the correct option

Q#2

option c is correct option when we solve an quadratic equation it gives two solution one is positive and other is negative as we know that income cannot be negative

hence only one viable solution exists when we solve this

y=4 $x^{2}$ +3x+45 quadratic equation

8 0

3 years ago

Read 2 more answers

Is Figure B a scale copy of Figure A?

r-ruslan [8.4K]

Answer:

No.

Step-by-step explanation:

A scale copy would see that both the x and y amount would be scaled proportionally. In this case, the x goes across 5 for figure A, and does the same for Figure B. On the other hand, the scaling for y is 2 for figure A, while it is 5 for figure B.

If it is a scale copy then.

If Figure A have the scaling of: 3, 5, 5, x, then Figure B, if given one scale of 5, should have

3:5

5:8.33

x:(x * ~1.67)

Therefore, the scaling is off.

6 0

3 years ago

Read 2 more answers

PLEASE HELP ME WITH MATH!!! Thank you

Lina20 [59]

Look at the 2 graphs. Restaurant A, the median is 35 (inside the box of plot). For Restaurant B, the median is 25 (shown as a line on that plot).
So 35-25=10 and therefore the median number of employees at Restaurant A is 10 more than the median at Restaurant B.

5 0

4 years ago

Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee

lubasha [3.4K]

Answer:

(a) 19.6 ms⁻¹

(b) 2 s

(d) 4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

$\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}$

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

$s=19.6 \quad v=0 \quad a=-9.8$

$\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}$

Therefore, the initial speed is 19.6 ms⁻¹.

Using the same values as for part (a):

$\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}$

Therefore, the time taken to reach the highest point is 2 seconds.

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

$u=19.6 \quad a=-9.8 \quad t=1$

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

$u=0 \quad a=9.8 \quad t=1$

(acceleration is now positive as we are taking ↓ as positive).

$\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}$

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

From part (a) we know that the time taken to reach the highest point is 2 seconds. Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0

1 year ago

Ryan is putting a clothesline in his rectangular backyard. He wants to put it between two trees on the edge of his property. He

Rama09 [41]

Answer:

25m

Explanation:

The first figure attached is the sketch of the property mentioned on the question, where all the lengths listed are in meters.

The second figure attached identifies a right triangle whose hypotenuse, c, is the smallest length that the clothesline can measure; and the legs measure 20m, and 20m - 10m = 15m.

Now, you can use the Pythagorean theorem to find the smallest length of the clothesline:

$c^2=(20m)^2+(15m)^2\\\\c^2=625m^2\\\\c=\sqrt{625m^2} \\\\c=25m\longleftarrow answer$

6 0

3 years ago

Read 2 more answers