Dimensions of a house are 500

Mathematics

1 answer:

Olin [163]3 years ago

5 0

Answer:

the answer is 20in. by 5in.

Step-by-step explanation:

you have to divide both of the numbers by 25

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Find the equation of the parabola whose vertex is the origin and whose directrix is x = -4

lapo4ka [179]

Try this option:

1) if V(0;0) and x= -4, then common view of the required equiation is:

(y-k)²=4p(x-h), where focus is in (h-p;k), the vertex is in (h;k), the directrix is x=h+p, p<0 and y=k is simmetry axis;

2) if the V(0;0), then h=k=0 and the required equiation is:

y²=4px;

3) if the directrix equation is x=h+p, where h=0, then p= -4 (according to the condition the directrix equation is x= -4), then the required equation is:

y²= -16x

answer: y²= -16x

6 0

3 years ago

From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends

valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let, AB = y, DC = x.

In Δ ABC,

$\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}$

Similarly, in Δ ACD,

$\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft$