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2 years ago

5

PLEASE HELP ASAP!! find the area of the shaded region in the figure below round to the nearest hundredth if necessary

2 answers:

Juliette [100K]2 years ago

5 0

Answer:

45cm^2

Step-by-step explanation:

Area of shaded region = area of outer triangle – area of the unshaded triangle

the area of a triangle is (base × height) ÷ 2

Area of outer triangle:

(12 × 5) ÷ 2 = 30cm^2

the area of the unshaded triangle:

(15 × 10) ÷ 2 = 75cm^2

area of shaded region = 75 - 30 = 45cm^2

Naily [24]2 years ago

3 0

I have solved the question
75cm^2

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3 years ago

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If f(x)=4x^2 and g(x)=x+1, find (fog)(x)

iren [92.7K]

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Step-by-step explanation:

Given: $f(x) = 4x^2 \text { and } g(x) = x+1$

To find: $(f_\circ g)(x)$

As we know it is composition function which means that g(x) function is in f(x) function.

So we have

$(f_\circ g) (x) = f[g(x)]$

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4 years ago

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3 years ago

Find the coordinates of the point 7/10 of the way from A to B. a=(-3,-6) b=(12,4)

Artemon [7]

Answer:

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

Step-by-step explanation:

Let be $A = (-3,-6)$ and $B = (12, 4)$ endpoints of segment AB and $M$ a point located 7/10 the way from A to B. Vectorially, we get this formula:

$\overrightarrow {AM} = \frac{7}{10}\cdot \overrightarrow {AB}$

$\vec M - \vec A = \frac{7}{10}\cdot (\vec B - \vec A)$

By Linear Algebra we get the location of $M$ :

$\vec M = \vec A + \frac{7}{10}\cdot (\vec B - \vec A)$

$\vec M = \vec A +\frac{7}{10}\cdot \vec B - \frac{7}{10}\cdot \vec A$

$\vec M = \frac{3}{10}\cdot \vec A + \frac{7}{10}\cdot \vec B$

If we know that $\vec A = (-3,-6)$ and $\vec B = (12, 4)$ , then:

$\vec M = \frac{3}{10}\cdot (-3,-6)+\frac{7}{10}\cdot (12,4)$

$\vec M = \left(-\frac{9}{10},-\frac{9}{5} \right)+\left(\frac{42}{5} ,\frac{14}{5} \right)$

$\vec M =\left(-\frac{9}{10}+\frac{42}{5} ,-\frac{9}{5}+\frac{14}{5} \right)$

$\vec M = \left(\frac{15}{2} ,1\right)$

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

6 0

3 years ago