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2 years ago

There are 8 girls and 5 boys on the math team. Which ratio correctly describes the relationship between boys and girls

Mathematics

2 answers:

Dimas [21]2 years ago

5 0

Answer:

8:5

Step-by-step explanation:

Andrej [43]2 years ago

5 0

8:5 because it’s girls to boys, and that’s the lowest you can simplify them

Sure hope this helps you. And good luck on whatever math test/exam is coming up. I'm surly know that you will do your very best. And pls mark me brainiest

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Fertilizer for your lawn has to be mixed with water. Every 5 ounces of mix needs 1 gallon of water. How many cups of mix do you

Mkey [24]

Answer:

5 cups

Step-by-step explanation:

There are 8 ounces in a cup, so the ratio given is ...

5/8 cup : 1 gallon

Multiplying this by 8, we have ...

5 cups : 8 gallons

If you are using 8 gallons of water, you need 5 cups of mix.

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3 years ago

Read 2 more answers

A bakery sells ,medium and large pies. The ratio small:medium:large is 3:5:2 .What fraction of the pies are large ?

SOVA2 [1]

2/8

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2 years ago

Suppose you are driving to visit a friend in another state. You are driving at an average rate of 50 miles per hour. You must dr

bagirrra123 [75]

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3 years ago

Complete the proof of the Chinese remainder theorem by showing that the simultaneous solution of a system of linear congruences

Semenov [28]

Answer:

Sorry I don`t know this

Step-by-step explanation:

I wish I could help you with it

7 0

3 years ago

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

sergeinik [125]

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$ , so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.

5 0

3 years ago