2 years ago

I will mark brain Can someone help me do not answer if u dont know

Mathematics

2 answers:

meriva2 years ago

6 0

Answer:

1. y = 1/3x+-1

2. y = 4

Hope this helps man :D

Otrada [13]2 years ago

4 0

I was about to answer until i saw someone already did ! good luck tho

You might be interested in

Students have a height is less than what number 58?62?64?70?

olya-2409 [2.1K]

Answer:

C 64 students have half the height

7 0

2 years ago

Helppp plssss this is due soooon

leva [86]

76;(/:14 volume of 6

7 0

2 years ago

Read 2 more answers

What is the sum of 4*8<br><img src="https://tex.z-dn.net/?f=4%20%5Ctimes%208" id="TexFormula1" title="4 \times 8" alt="4 \times

Inessa [10]

See picture reply for answer.

5 0

2 years ago

Read 2 more answers

The length and width of a rectangle are measured as 55 cm and 49 cm, respectively, with an error in measurement of at most 0.1 c

valentina_108 [34]

Answer:

The maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

Step-by-step explanation:

The area of a rectangle with length $L$ and width $W$ is $A= L\cdot W$ so the differential of <em>A</em> is

$dA=\frac{\partial A}{\partial L} \Delta L+\frac{\partial A}{\partial W} \Delta W$

$\frac{\partial A}{\partial L} = W\\\frac{\partial A}{\partial W}=L$ so

$dA=W\Delta L+L \Delta W$

We know that each error is at most 0.1 cm, we have $|\Delta L|\leq 0.1$ , $|\Delta W|\leq 0.1$ . To find the maximum error in the calculated area of the rectangle we take $\Delta L = 0.1$ , $\Delta W = 0.1$ and $L=55$ , $W=49$ . This gives

$dA=49\cdot 0.1+55 \cdot 0.1$

$dA=10.4$

Thus the maximum error in the calculated area of the rectangle is $10.4 \:cm^2$

4 0

2 years ago

HELP I NEED HELP ASAP

AfilCa [17]

Answer :
x = -1 and y = -4

Solution :

4 0

2 years ago

Read 2 more answers