All categories

3 years ago

Plz, answer this it is really easy.

Mathematics

1 answer:

dimulka [17.4K]3 years ago

7 0

Answer:

v = 17.4 meters

Step-by-step explanation:

This is a triangle with height (red dotted line) given.

The area of a triangle is given by the formula:

$A=\frac{1}{2}bh$

Where

A is the area

b is the base length

h is the height

Given height = 12, and

area = 104.4

THe base is labeled as "v", so we find v by replacing into formula:

$A=\frac{1}{2}bh\\104.4=\frac{1}{2}(v)(12)\\104.4=6v\\v=\frac{104.4}{6}\\v=17.4$

The base, v, is 17.4 meters

You might be interested in

Write an equation of the line that passes through (4,

Dmitry [639]

Answer:

Step-by-step explanation:

y = x + 5

y - 6 = x - 4

y = x + 2

8 0

4 years ago

Chloe is installing new windows in her house and needs to determine the amount of sealant she will need around the outside edge

Bingel [31]

Perimeter of window is 192 inches.

Step-by-step explanation:

We are given:

Area of rectangular window = 2160 inches square

Length of rectangular window = 60 inches

Required:

Perimeter of window=?

Formula to find Perimeter is:

$Perimeter=2(length+Width)$

We first find width of rectangular window by using formula of area:

$Area=Length*Width$

Putting values:

$Area=Length*Width\\2160=60*Width\\Width=\frac{2160}{60}\\Width=36\,\,inches$

So, width = 36 inches

Now finding perimeter:

$Perimeter=2(length+Width)\\Perimeter=2(60+36)\\Perimeter=2(96)\\Perimeter=192\,\,inches$

So, perimeter of window is 192 inches.

Keywords: Perimeter of rectangle

Learn more about Perimeter of rectangle at:

#learnwithBrainly

3 0

3 years ago

One fourth the difference of four sixths and three twelfth

Lyrx [107]

1/4 (4/6 - 3/12) = 1/4 (5/12) = 5/48

3 0

4 years ago

....need help on this

Leona [35]

Answer:

ASA statement

Step-by-step explanation:

In quadrilateral ABCD, diagonal AC bisects $\angle BAD$ and $\angle BCD$ (given)

$In\: \triangle ABC \: and \: \triangle ADC$

$\angle CAB\cong\angle CAD$

(By angle bisector property)

$AC\cong AC.... (common\: side)$

$\angle ACB\cong\angle ACD$

(By angle bisector property)

$\therefore \triangle ABC \cong \triangle ADC$

(By ASA Statement)

3 0

3 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72

$x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37$

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0

3 years ago