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2 years ago

9

a section of a stained glass window is shaped like a parallelogram. Its base is 6.5 inches, and its height is 4 inches. how much

glass is needed to cover the section completely?

1 answer:

Bond [772]2 years ago

8 0

Area is base multiplied by height, so to get the answer you'd need to multiply 6.5 and 4

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Graph the function f(x)= 2x -4

galben [10]

The y intercept and -4 and going down 2 and 1 right u til you reach the end of the graph

7 0

1 year ago

This graph shows a proportional relationship.<br> What is the constant of proportionality?

krek1111 [17]

hookes law gives the equation F=kx where F is the elastic force and k is the constant and x or small e is extension if we draw a graph you'll see that the graph increases by the same ratio every single time hence giving a straight line show that they are F and x are propotional to a certain limit

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2 years ago

Read 2 more answers

There were 8 red cars and 12 blue cars in the parking lot what is the ratio of the number of red cars to number of blue cars

salantis [7]

Answer:

2 : 3

Step-by-step explanation:

<em>Hello fellow human!</em>

red : blue - 8 : 12

The ratio is divisible by 4 - (8/4) = 2 ; (12/4) = 3

Hence, 2 : 3

8 0

2 years ago

A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta

nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is $\mu=30$

Standard deviation of Eskimo dogs is $\sigma=2$

The range of weights would 99.7% of the dogs have,

$R=\mu\pm3\sigma$

$R=30\pm3(2)$

$R=30\pm6$

$R=30+6,30-6$

$R=36,24$

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

2 years ago