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borishaifa [10]
3 years ago
12

Veronica copies her uncle's address and phone number into her contact list. His area code is 775. His ZIP code is 89507. Which s

tatement about the value of the 5 in 775 and 89,507 is true? * 5 points It is the same in both numbers It is 10 times as great in the ZIP code than it is in the area code It is 100 times as great in the ZIP code than it is in the area code It is 10 times as great in the area code than it is in the ZIP code
Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer:It is 100 times as great in the ZIP code than it is in the area code

Step-by-step explanation:

step 1

Veronica 's uncle ZiP CODE = 89507 = The value of 5 here, represents 5 Hundred-----500

Veronica 's uncle Area code= 775 = The value of 5 here  represents 5 units--5

it will take a 100 times the value of 5 in the area code to be same as the value of 5 in the ZIP code

Step 2:

Therefore, we can say that the value of 5 is 100 times as great as the ZIP code than it is in the area code.

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Please help please ​
jarptica [38.1K]

Answer:

1.) supplementary  m<1+m<2=180 and m<2+m<3=180

2.) congruent, supplementary

Step-by-step explanation:

the answers are in this order above

5 0
3 years ago
Did I do my math homework right?
anastassius [24]
Yes you did it correctly but change problem 5’s denominator on the right to a 6
7 0
3 years ago
Find the volume of each figure. Round your answers to the nearest hundredth, if necessary
SCORPION-xisa [38]

Answer:

666.67π .ft^3

<em>here's</em><em> your</em><em> solution</em>

<em>=</em><em>></em><em> </em><em>radius</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>1</em><em>0</em><em> </em><em>ft</em>

<em>=</em><em>></em><em> </em><em>height</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>2</em><em>0</em><em> </em><em>ft</em>

<em>=</em><em>></em><em> </em><em>volume</em><em> of</em><em> </em><em>cone </em><em>=</em><em> </em><em>πr^</em><em>2</em><em>h</em><em>/</em><em>3</em>

<em>=</em><em>></em><em> </em><em>putting</em><em> the</em><em> value</em><em> of</em><em> </em><em>radius</em><em> and</em><em> height</em><em> </em>

<em>=</em><em>></em><em> </em><em> </em><em> </em><em>volume</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>^</em><em>2</em><em>*</em><em>2</em><em>0</em><em>/</em><em>3</em><em>π</em>

<em>=</em><em>></em><em> </em><em>volume</em><em> </em><em>=</em><em> </em><em>6</em><em>6</em><em>6</em><em>.</em><em>6</em><em>7</em><em>π</em><em> </em><em>.</em><em>ft^</em><em>3</em>

<em>hope</em><em> it</em><em> helps</em>

5 0
2 years ago
Consider the figure.<br> What is JL?
lisov135 [29]

The answer for the above mentioned problem is JL = 12.5

Step by step explanation:

Given:

JM = 8

KM = 6

To Find:

JL = ?

Formula to be used:

KM^2 = JM x ML

In order to find " JL" we must first find "ML",

KM^2 = JM x ML

         6^2 = 8 x ML

         36 = 8 x ML

         36/8 = ML

         ML = 4.5

Now JL = 4.5+8

             = 12.5

Thus the value of JL = 12.5

3 0
3 years ago
Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
Ipatiy [6.2K]

Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

(C) y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}

(D) y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R},

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

(D) Using C, let be y=me^{3t} we obtain that it must satisfies 3^2m-4m=1\Rightarrow m=1/5 and then the general solution is y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5 with k_1,k_2\in\mathbb{R}, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).  

4 0
3 years ago
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