All categories

3 years ago

What’s the answer to this problem

Mathematics

1 answer:

KonstantinChe [14]3 years ago

3 0

Fivuvucucvuvhgjkdjxjxgtcycuc

You might be interested in

At the annual dogs show chantel noticed that there were three more Scottie’s than schnauzers. She also realized that the number

galben [10]

<span>Let us start with the schnauzers, the easiest one to imagine. Let us assume that there were x number of schnauzers. Scottie's are 3 more than schnauzers. So their number is x+3 Wire haired terriers are 5 less than twice the number of schnauzers. So their number is 2x -5 (2x for twice the number of schnauzers) Now add all these numbers. That is x + x+3 + 2x-5 = 4x -2 The total number of dogs, 78, is given in the question Now we know that 4x -2 =78 4x = 78 +2 = 80 Therefore x = 80/4 = 20. So there were 20 schnauzers, 23 Scottieâ€™s and 35 wire haired terriers</span>

6 0

3 years ago

Did I get this correct

saveliy_v [14]

Answer:

Yep

Step-by-step explanation:

7 0

3 years ago

PLEASE HELP!!!! WORTH 20 points it’s all I have show all of your work!!!!!!!

stiv31 [10]

Answer:

Part A: [0,2], Part B: [2, 4], Part C: [8,10], Part D: 0 ft

Step-by-step explanation:

Part A: The time between 0 seconds to 2 seconds is the time when it's increasing. There is no other time like that.

Part B: The graph shows a straight line so therefore it is staying the same.

Part C: The graph shows a sharp decline.

Part D: Starting from the 10th second, the balloon already hit to ground so it would only make sense for it to stay there...

8 0

3 years ago

Tom counted the number of cards in his collection for 10 weeks. By week 5, he had collected 12 cards. This graph shows this info

Karolina [17]

Answer:

i think it would be 17

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Evaluate the integral ∫2−1|x−1|dx

defon

I think you might be referring to the definite integral,

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx$

Recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

Then $|x-1|=x-1$ if $x\ge1$ , and $|x-1|=1-x$ is $x$ . So spliting up the integral at <em>x</em> = 1, we have

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \int_{-1}^1(1-x)\,\mathrm dx + \int_1^2(x-1)\,\mathrm dx$

The rest is simple:

$\displaystyle \int_{-1}^2|x-1|\,\mathrm dx = \left(x-\dfrac{x^2}2\right)\bigg|_{-1}^1 + \left(\dfrac{x^2}2-x\right)\bigg|_1^2 \\\\ = \left(\left(1-\frac12\right)-\left(-1-\frac12\right)\right) + \left(\left(2-2\right)-\left(\frac12-1\right)\right) \\\\ = \boxed{\frac52}$

5 0

3 years ago