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3 years ago

9

A hockey team is revealing their championship banner outside their stadium. The banner is in the shape of a triangle with a heig

ht that is 4 times as tall as the base length of the banner. If the total area of the banner is 18 square meters (18m2), what is the length of the base of the banner in meters?

1 answer:

kipiarov [429]3 years ago

5 0

Answer: the length of the base of the banner is 3 meters.

Step-by-step explanation:

Assuming the banner is in the shape of a right angle triangle, we would apply the formula for determining the area of a triangle which is expressed as

Area = 1/2 × base × height

Let b represent the base

The banner is in the shape of a triangle with a height that is 4 times as tall as the base length of the banner. It means that the height of the banner is 4b

If the total area of the banner is 18 square meters, it means that

18 = 1/2 × b × 4b

18 = 2b²

b² = 18/2 = 9

b = √9 = 3 meters

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Es posible dibujar 9 segmentos de línea de manera que cada segmento interseque exactamente 1 de los otros segmentos?

uysha [10]

Answer: No es posible (It is not possible)

Step-by-step explanation:

The question in english is as follows:

Is it possible to draw 9 line segments so that each segment crosses exactly 1 of the other segments?

A necessary condition for two lines to intersect is that they must be in the same plane (they must be coplanar), being their intersection a single point.

Now, if we want several lines to intersect only once, we need to have an <u>even number</u> of line segments. However, this is not the case because <u>9 is odd. </u>

Therefore, it is not possible.

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3 years ago

Evaluate (2-5i)(p+q)(i) when p= 2 and q=5i

Mumz [18]

Answer:

it's 25qr 78 5×4 and then 12÷5= 2r28

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3 years ago

If θ is an angle in standard position and its terminal side passes through the point (-4,-5), find the exact value of

NARA [144]

9514 1404 393

Answer:

cot(θ) = 4/5

Step-by-step explanation:

In the polar/rectangular coordinate representation (x, y) ⇔ (r; θ), we know that ...

(x, y) = (r·cos(θ), r·sin(θ))

From the various trig definitions and identities, we also know that ...

cot(θ) = cos(θ)/sin(θ) = (x/r)/(y/r) = x/y

For the given (x, y) = (-4, -5), the cotangent is ...

cot(θ) = -4/-5 = 4/5

7 0

3 years ago

Someone please help me

Lady_Fox [76]

Answer:

40% of 80=32

25% of 60=15

10% of 90=9

50% of 70=35

80% of 500=400

75% of 80=60

90% of 250=225

65% of 400=260

85% of 800=680

55% of 140=77

45% of 160=72

95% of 180= 171

70% of 720=504

15% of 220=33

65% of 200= 130

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Use the Normal model N(1133,78) for the weights of steers.

Rudiy27

Answer:

Step-by-step explanation:

Let $X\sim N(1133, 78)$ .

Here, the mean is 1133 and standard deviation is 78.

Since we don't have the table for N(1133, 78), so we use the standard table for N(0,1),

a)

All we have to do is find, within the table the specific percentile.

now to find (42nd percentile) 0.42 in the normal table N(0,1) and extract the number on the row and column.

$P(X\leq x)=0.42$

Using: $Z=\frac{X-\mu}{\sigma}$ where $\mu$ is the mean and $\sigma$ is the standard deviation.

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.42$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.42$

Now, using Standard Normal table to find the value of z- score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z.

i.e, $z= -0.20$

Now, putting this value in $z=\frac{x-1133}{78}$ , to find x;

$\frac{x-1133}{78}=-0.20$

On Simplify, we get;

$x=1,117.4$ pounds

b)

Similarly, find the weight for 91st percentile.

Follow the same steps that we have done in part (a),

$P(X\leq x)=0.91$ or

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.91$

⇒ $P(Z \leq \frac{x-1133}{78})=0.91$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.91$

Now, use normal table value to find the z-score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z

we have, z= 1.34

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=1.34$

On simplify, we get

x= 1,237.52 pounds

c)

the interquartile range (IQR)=3rd Quartile - 1st quartile

First find the 1st quartile:

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.25$

⇒ $P(Z \leq \frac{x-1133}{78})=0.25$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.25$

Now, we use normal table to find that z=-0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=-0.675$

On simplify, we get

x= 1,080.35 pounds

Similarly, for third quartile

By symmetry z=0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=0.675$

On simplify, we get

x= 1,185.65 pounds

then, IQR = 1185.65 -1080.35=105.3 pounds

7 0

3 years ago