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4 years ago

What is the partial product for 42x25

Mathematics

1 answer:

Hoochie [10]4 years ago

3 0

I think it will be 1,050 because I multiply and that is what I got for my answer

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Find the area of the shaded region

Brilliant_brown [7]

Answer:

If you multiply the length and width it's already more than D and b so it's between 41 and 71 , I believe it's 71 though

4 0

3 years ago

Terms are also expressions. No Some А All PLZ ANSWER FAST

luda_lava [24]

The answer would be some

3 0

3 years ago

The high school took two field trips to the Newport Aquarium many many years ago. They used school buses and vans to transport a

il63 [147K]

Answer:

bus = 43

van 12

Step-by-step explanation:

This can be solved using simultaneous equations

Let v represent the number of students that a van carries

Let b represent the number of students that a bus carries

the following equations can be derived from the question

3v + 2b = 122 eqn 1

5v + 3b = 189 eqn 2

Multiply eqn 1 by 5 and eqn 2 by 3

15v + 10b = 610 eqn 3

15v + 9b = 567 eqn 4

Subtract equation 4 from 3

b = 43

Substitute for b in equation 1

3v + 2(43) = 122

solve for v

v = 12

5 0

3 years ago

Can Somebody Please Help Me With This!!

adelina 88 [10]

Answer:

I think A

Step-by-step explanation:

sorry if it is not correct hope it helps

8 0

3 years ago

Read 2 more answers

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

3 years ago