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3 years ago

The distance between two towns on a map is 6cm. The actual distance between two towns is 48 km. What is the scale on the map?

Mathematics

1 answer:

iragen [17]3 years ago

7 0

Step-by-step explanation:

6cm : 48km

=> 1cm : 48km * (1/6) = 1cm : 8km. (C)

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The cat’s weight changed -8 oz. while she was sick. Which of the following shows a greater change in weight? A. Loss of 9 oz B.

HACTEHA [7]

<u>Answer:</u>

The correct answer option is A. Loss of 9 oz.

<u>Step-by-step explanation:</u>

We are given that a cat's weight change -8 oz. while she was sick. It means that the cat lost 8 ounces of weight.

We are to determine whether which of the given answer options show a greater change in weight.

The correct answer for this is: loss of 9 oz which is a greater loss than 8 oz.

4 0

3 years ago

Solve the initial value problems.

slavikrds [6]

Both equations are linear, so I'll use the integrating factor method.

The first ODE

$xy' + (x+1)y = 0 \implies y' + \dfrac{x+1}x y = 0$

has integrating factor

$\exp\left(\displaystyle \int\frac{x+1}x \, dx\right) =\exp\left(x+\ln(x)\right) = xe^x$

In the original equation, multiply both sides by eˣ :

$xe^x y' + (x+1) e^x y = 0$

Observe that

d/dx [xeˣ] = eˣ + xeˣ = (x + 1) eˣ

so that the left side is the derivative of a product, namely

$\left(xe^xy\right)' = 0$

Integrate both sides with respect to x :

$\displaystyle \int \left(xe^xy\right)' \, dx = \int 0 \, dx$

$xe^xy = C$

Solve for y :

$y = \dfrac{C}{xe^x}$

Use the given initial condition to solve for C. When x = 1, y = 2, so

$2 = \dfrac{C}{1\cdot e^1} \implies C = 2e$

Then the particular solution is

$\boxed{y = \dfrac{2e}{xe^x} = \dfrac{2e^{1-x}}x}$

The second ODE

$(1+x^2)y' - 2xy = 0 \implies y' - \dfrac{2x}{1+x^2} y = 0$

has integrating factor

$\exp\left(\displaystyle \int -\frac{2x}{1+x^2} \, dx\right) = \exp\left(-\ln(1+x^2)\right) = \dfrac1{1+x^2}$

Multiply both sides of the equation by 1/(1 + x²) :

$\dfrac1{1+x^2} y' - \dfrac{2x}{(1+x^2)^2} y = 0$

and observe that

d/dx[1/(1 + x²)] = -2x/(1 + x²)²

Then

$\left(\dfrac1{1+x^2}y\right)' = 0$

$\dfrac1{1+x^2}y = C$

$y = C(1 + x^2)$

When x = 0, y = 3, so

$3 = C(1+0^2) \implies C=3$

$\implies \boxed{y = 3(1 + x^2) = 3 + 3x^2}$

7 0

2 years ago

What are the zeros of the polynomial function f(x)=x^2-5x-6?

lora16 [44]

$f(x) = \: {x}^{2} - 5x - 6 \\ \\ f(x) = \: {x}^{2} - 6x + x - 6 \\ \\ f(x) = \: (x - 6)(x + 1) = \: 0 \\ \\ || \: x = \: 6 || \: \: \: \: or \: \: \: || x = - 1 || \: \: \: \: \: Ans.$

Hope it helps you :)

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4 years ago

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Fynjy0 [20]

There are 3 sides and 3 angles per shape, because they are both triangles. hope this helped you out!!!

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3 years ago

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Mike says the opposite of 5 is –5 and the absolute value of 5 is –5. Is he correct? Use the drop-down menus to explain your answ

kolbaska11 [484]

No cs the absolute value is how far it is from zero and -5 is 5 spots from zero so he's incorrect

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3 years ago