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3 years ago

What is the solution set for 4x + 12 ≤ 8?

Mathematics

2 answers:

timurjin [86]3 years ago

7 0

Answer:x

≥

−

Step-by-step explanation:

aivan3 [116]3 years ago

5 0

Answer:

Inequality Form: x ≤ -1

Interval Notation: (−∞,−1]

You might be interested in

How much warmer is 12°C than -20°C?

zvonat [6]

To put it simply, 12 degrees C is 12 away from 0 and -20 degrees C is 20 away from 0. Taking these two inputs and adding them together should give you 32. So 12 degrees C is 32 degrees warmer than -20 degrees C

8 0

3 years ago

Read 2 more answers

If it cost me 375.00 to stay in a hotel for 3 days. How much it cost for one day.

marusya05 [52]

Answer:

125

Step-by-step explanation:

8 0

3 years ago

Solve surface area please! Quick it’s due soon

just olya [345]

Answer:

SA = 470 m²

volume = 494 m³

Step-by-step explanation:

<u>Individual areas</u>

Back = 4 x 19 = 76 m²

Top = 5 x 19 = 95 m²

sloped side = 5 x 19 = 95 m²

base = 8 x 19 = 152 m²

end = (4 x 5) + 1/2(3 x 4) = 26 m²

Total SA = 76 + 95 + 95 + 152 + (2 x 26) = 470 m²

<u>Volume</u>

volume = (4 x 5 x 19) + 1/2(3 x 4 x 19) = 494 m³

5 0

2 years ago

By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00pm positive or negative?

Ad libitum [116K]

Answer:

Part a) The temperature at noon was -3°F

Part b) The temperature at 3:00 pm was +2°F

Part c) The temperature at 11:00 pm was -6°F

Step-by-step explanation:

The complete question is

When Leo woke up he saw that the temperature was -8°F. By noon the temperature has increased 5°F. Part a) What was the temperature at noon?

Part b) By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00 pm positive or negative?

Part c) By 11:00 pm, the temperature had dropped 8°F. Was the

temperature at 11:00 PM positive or negative? Explain.

Part a) we know that

You can use a number line to adds the numbers

-8+5=-3°F

therefore

The temperature at noon was -3°F

Part b) we know that

the temperature increased by another 5 degrees

You can use a number line to adds the numbers

-3+5=+2°F

therefore

The temperature at 3:00 pm was +2°F

Part c) we know that

the temperature had dropped 8°F

You can use a number line to adds the numbers

2-8=-6°F

therefore

The temperature at 11:00 pm was -6°F

6 0

3 years ago

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago