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lesya [120]
2 years ago
5

6x-2y=48 Find the y and x intercept Write in ordered Pairs

Mathematics
2 answers:
Andrej [43]2 years ago
8 0

Answer:

Correct you are correct

Step-by-step explanation:

maksim [4K]2 years ago
4 0
Y intercept (0,-24)
X intercept (0,8)
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Can someone answer part B for me? Thank you and will give brainliest.
frez [133]

Answer:

The answer to part B is B.

Step-by-step explanation:

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4 0
3 years ago
If Joseph buys a T.V. At 35% off and it’s original costs $599.99 how much did he pay for the tv?
Monica [59]

Answer: He paid \$389.9935 for the T.V.

Step-by-step explanation:

We need to analize the information provided:

- The original cost of the T.V. was $599.99

- Joseph buys it at 35% off the original cost.

Let be "x" the amount Joseph paid for the T.V.

We can calculate "x" with the following expression:

x=\$599.99-(\$599.99)(0.35)\\\\x=\$389.9935

Therefore, he paid \$389.9935 for the T.V.

7 0
3 years ago
Find the percent: 80% of 8
8090 [49]

Answer:

6.4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the
Vedmedyk [2.9K]
(a) Using Newton's Law of Cooling, \dfrac{dT}{dt} = k(T - T_s), we have \dfrac{dT}{dt} = k(T - 75) where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get \dfrac{dT}{T - 75} = k dt. Integrate both sides to get \ln|T - 75| = kt + C.

Since T(0) = 185, we solve for C:
|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110
So we get \ln|T - 75| = kt + \ln 110. Use T(30) = 150 to solve for k:
\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)

So

\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t}  \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}

But choose Positive because T > 75. Temp of turkey can't go under.

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)}  = 136.929 \approx 137{}^{\circ}F

(b)

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30}  \\
100 = 75 + 110(15/22)^{t/30}   \\
25 = 110(15/22)^{t/30}  
\frac{25}{110} = (15/22)^{t/30}   \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22)  \approx 116\ \mathrm{min}

Dogs of the AMS.
4 0
3 years ago
A formula for a function y=f(x) is f(x)=x^2-10x,xless than or equal 5. Find f^-1(x) and indentify the domain and range of f^-1(x
evablogger [386]
F(x) =x² - 10x,  f⁻¹(x) =?

1st find the missing square of x²-10x, ==> (x-5)² - 25

y= (x-5)² - 25; replace x by y and v0ce versa: x= (y-5)² -25
or
(y-5)² = x+25
y-5 = √(x+25) and y = √(x+25) -5

Domain = {x∈R: X>= 25} AND Range ={y∈R: y>= 5}

6 0
3 years ago
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