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3 years ago

10

Gordon recorded a temperature of -22°C, and Maki recorded a temperature of 13°C. How much warmer was the temperature that Maki r

ecorded?

1 answer:

Nikolay [14]3 years ago

5 0

Answer:

35°C

Step-by-step explanation:

13+22=35

You might be interested in

A number sentence is shown below:

d1i1m1o1n [39]

Number sentence is simply a sentence, that uses numbers and arithmetic signs.

The context that can be created from the sentence is: <em>How many bits are there, in </em> $32\frac 14$ bytes
The verbal meaning of $32\frac 14 \div \frac 18 = 258$ is: <em>When </em> $32\frac 14$ <em> is divided by </em> $\frac 18$ <em>, the result is </em> $258$ <em />

<em />

Given

$32\frac 14 \div \frac 18 = 258$

<u />

<u>(a) Context or story</u>

A context is as follows:

<em>There are 8 bits in a byte; How many bits are there, in </em> $32\frac 14$ bytes

<u />

<u>(b) As a multiplication</u>

$32\frac 14 \div \frac 18 = 258$

Change the <em>division to multiplication </em>as follows:

$32\frac 14 \times \frac 81 = 258$

<u />

<u>(c) Verbal explanation how the numbers are related</u>

When $32\frac 14$ is divided by $\frac 18$ , the result is $258$

Read more about number sentence at:

brainly.com/question/17322770

4 0

2 years ago

Please help with this worksheet I don’t know any of it

sammy [17]

Answer:

4. 1/25

5. 1/16

6. 2/9

7. 5/567

8. 1/40

9. 2/175

10. 1/30

11. 2/5

12. 9/8

13. 4/3

14. 1

Step-by-step explanation:

sorry I only know how to do 4-14. I hope this helps you.

4 0

3 years ago

When using a debit card: Question options: A:There are always fees. b:Identification is always required. c:You still need to rec

DIA [1.3K]

Answer:

D.

Step-by-step explanation:

Since this isn't a credit card, there are no interests or fees on a debit card, so A is incorrect. B is also incorrect. You only need identification when you are withdrawing or depositing at a bank, but purchases made in stores or online do not need your identification. You also don't need to record transactions in your checkbook (but it is recommend to keep track of purchases). Modern day technology already records transaction history and all you need to do is access it online.

D is correct because if someone steals your PIN for your debit card, they could go to stores and use that money. You can dispute charges and report to the bank if that happens.

3 0

3 years ago

Someone who knows the remainder theorem help!

Damm [24]

Answer:

○ D. Yes, x = 12 is a zero of the polynomial.

The quotient is x + 22, and the remainder is 0.

Step-by-step explanation:

On a second thought, I knew something similar to that theorem because factoring them would determine if it has a remainder:

[x - 12][x + 22]

I am joyous to assist you anytime.

* I apologize for the previous answer I gave you.

5 0

3 years ago

Suppose that \nabla f(x,y,z) = 2xyze^{x^2}\mathbf{i} + ze^{x^2}\mathbf{j} + ye^{x^2}\mathbf{k}. if f(0,0,0) = 2, find f(1,1,1).

lesya [120]

The simplest path from (0, 0, 0) to (1, 1, 1) is a straight line, denoted $C$ , which we can parameterize by the vector-valued function,

$\mathbf r(t)=(1-t)(\mathbf i+\mathbf j+\mathbf k)$

for $0\le t\le1$ , which has differential

$\mathrm d\mathbf r=-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

Then with $x(t)=y(t)=z(t)=1-t$ , we have

$\displaystyle\int_{\mathcal C}\nabla f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\nabla f(x(t),y(t),z(t))\cdot\mathrm d\mathbf r$

$=\displaystyle\int_{t=0}^{t=1}\left(2(1-t)^3e^{(1-t)^2}\,\mathbf i+(1-t)e^{(1-t)^2}\,\mathbf j+(1-t)e^{(1-t)^2}\,\mathbf k\right)\cdot-(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dt$

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)(t^2-2t+2)\,\mathrm dt$

Complete the square in the quadratic term of the integrand: $t^2-2t+2=(t-1)^2+1=(1-t)^2+1$ , then in the integral we substitute $u=1-t$ :

$\displaystyle=-2\int_{t=0}^{t=1}e^{(1-t)^2}(1-t)((1-t)^2+1)\,\mathrm dt$

$\displaystyle=-2\int_{u=0}^{u=1}e^{u^2}u(u^2+1)\,\mathrm du$

Make another substitution of $v=u^2$ :

$\displaystyle=-\int_{v=0}^{v=1}e^v(v+1)\,\mathrm dv$

Integrate by parts, taking

$r=v+1\implies\mathrm dr=\mathrm dv$

$\mathrm ds=e^v\,\mathrm dv\implies s=e^v$

$\displaystyle=-e^v(v+1)\bigg|_{v=0}^{v=1}+\int_{v=0}^{v=1}e^v\,\mathrm dv$

$\displaystyle=-(2e-1)+(e-1)=-e$

So, we have by the fundamental theorem of calculus that

$\displaystyle\int_C\nabla f(x,y,z)\cdot\mathrm d\mathbf r=f(1,1,1)-f(0,0,0)$

$\implies-e=f(1,1,1)-2$

$\implies f(1,1,1)=2-e$

3 0

3 years ago