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3 years ago

Help me please I beg

Mathematics

1 answer:

Readme [11.4K]3 years ago

4 0

As far as I understand A)

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3х – 8y = 34 7х + 4у = -34 Solve by elimination Y and x Please show steps ):

garri49 [273]

Answer:

The answer would be (-2, -5)

Step-by-step explanation:

3x - 8y= 34

2(7x + 4y = -34)

3x - 8y=34

14x + 8y= -68

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

17x= -34 divided by 17

x = -2

3(-2) -8y= 34

-6 -8y =34 add 6

-8y= 40 divide by -8

y = -5

if you want to check the answer replacing -2 for x and -5 for y

7(-2) + 4(-5)= -34

-14 + (-20) = -34

-34 = -34

I hope this help sorry I'm not good explaining

4 0

3 years ago

Can someone please help with this it’s worth 19 points?

nexus9112 [7]

Answer:

54, 36

Step-by-step explanation:

67+59+x=60*3

126+x=180

x=54

54 cones on Sunday

_____________________________________

180+x=54*4

180+x=216

x=36

36 cones on Monday

6 0

3 years ago

Help plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

sertanlavr [38]

y=1/2x the first bubble

4 0

3 years ago

Read 2 more answers

Need help pls pls pls pls pls pls

Crazy boy [7]

Answer:

Step-by-step explanation:

hi I think its d, sorry if I'm wrong

6 0

3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: Rate of change measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

3 years ago