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Anarel [89]
2 years ago
11

City R has a temperature of −2 °F. City S has a temperature of −6 °F. Use the number line shown to answer the questions: Number

line from negative 8 to positive 8 in increments of 1 is shown. Part A: Write an inequality to compare the temperatures of the two cities. (3 points) Part B: Explain what the inequality means in relation to the positions of these numbers on the number line. (4 points) Part C: Use the number line to explain which city is warmer. (3 points)
Mathematics
2 answers:
marta [7]2 years ago
6 0

Answer:

part c city r is warmer

Step-by-step explanation:

irina [24]2 years ago
3 0

Answer:warmeerrrr

Step-by-step explanation:

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Plz help I will give brainliest!
elena55 [62]

Answer:

b, 1 5/8

Step-by-step explanation:

the area is 4 15/32 square meters, and you find the area by doing length x width. the length is provided, 2 3/4. so all you have to do is divide 4 15/32 by 2 3/4 and you get 1 5/8

6 0
3 years ago
Read 2 more answers
The value of y varies directly with x. If x = 7, then y = 15.
Alona [7]

Answer:

5

Step-by-step explanation:

mark me brainliest plz

8 0
3 years ago
Read 2 more answers
Please help i need this done QUICK im timed ill give the best answer brainliest
seraphim [82]

Answer:

x=−32/5

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

1/2(1/4x−3/5)=1/4(2/5+3/4x)

(1/2)(1/4x)+(1/2)(−3/5)=(1/4)(2/5)+(1/4)(3/4x)(Distribute)(1/2)(1/4x)+(1/2)(−3/5)=(1/4)(2/5)+(1/4)(3/4x)(Distribute)

1/8x+−3/10=1/10+3/16x

1/8x+−3/10=3/16x+1/10

Step 2: Subtract 3/16x from both sides.

1/8x+−3/10−3/16x=3/16x+1/10−3/16x

−1/16x+−3/10=1/10

Step 3: Add 3/10 to both sides.

−1/16x+−3/10+3/10=1/10+3/10

−1/16x=2/5

Step 4: Multiply both sides by 16/(-1).

(16/−1)*(−1/16x)=(16/−1)*(2/5)

x=−3/25

7 0
3 years ago
Hey i rlly want to be done with thiss lol don't delete;-;
e-lub [12.9K]

Answer:

hello the correct answer is y=30x+20

Step-by-step explanation:

6 0
2 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
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