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2 years ago

Find the area of the triangle.

Mathematics

1 answer:

Step2247 [10]2 years ago

8 0

Answer:

$10units^2$

Step-by-step explanation:

Unshaded Right Triangle

$A=\frac{ab}{2} \\=\frac{4*2}{2} \\=4units^2$

Whole Triangle

$A=\frac{ab}{2} \\=\frac{4*7}{2}\\ =14units^2$

Shaded triangle

$14-4=10units^2$

If you dont want as many steps, just calculate the triangle and dont worry about the right triangle the formula is

$A=\frac{h_{b}b}{2} \\=\frac{4*5}{2}\\ =10units^2$

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In the first event, the eighth graders are running a baton relay race with three other classmates. The team's top speed for each

Vladimir [108]

Answer:

233.53 s

Step-by-step explanation:

The sum of best times for the legs is ...

56.81 +59.22 +57.39 +60.11 = 233.53 . . . seconds

Without additional information about the effects of training or the normal variation in times, this is the best prediction we can make.

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3 years ago

What is a reasonable estimate for the solution? O (1,-3/4) O (-3/4,1) O (-1,3/4) O (3/4,-1)

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Answer:

See Explanation

Step-by-step explanation:

Your question is incomplete, as the equations or graph or table(s) were not given.

However, I'll give a general way of solving this.

Take for instance, the equations are:

$y = \frac{4}{3}x - 1$

$y = \frac{2}{3}x - \frac{1}{2}$

To do this, we start by equating both equations.

$y = y$

i.e.

$\frac{4}{3}x - 1= \frac{2}{3}x - \frac{1}{2}$

Collect Like Terms

$\frac{4}{3}x - \frac{2}{3}x= 1 - \frac{1}{2}$

Take LCM

$\frac{4x- 2x}{3}= \frac{2 - 1}{2}$

$\frac{2x}{3}= \frac{1}{2}$

Cross Multiply

$2x * 2 = 3 * 1$

$4x = 3$

Make x the subject

$x = \frac{3}{4}$

Substitute 3/4 for x in $y = \frac{4}{3}x - 1$

$y = \frac{4}{3} * \frac{3}{4} - 1$

$y = 1 - 1$

$y = 0$

Hence:

$(x,y) = (\frac{3}{4},0)$

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2 years ago

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

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2 years ago

In a test of a gender-selection technique, results consisted of 248 baby girls and 13 baby boys. based on this result, what is

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2 years ago

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