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2 years ago

PLZ HELP QUICK!!

Mathematics

2 answers:

Julli [10]2 years ago

8 0

Answer: A(y=x/2)
Direct variation describes the relationship between the two variables

Murljashka [212]2 years ago

5 0

The answer would be A

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Geometry Homework Help Please?

kolbaska11 [484]

Ok what is the geometry u need help with

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3 years ago

a rectangular sign has a length of 3 and 1/2 ft and width of 1 and 1/8 ft will the area of the sign be greater or less than 3 an

svetoff [14.1K]

Answer:

The area is 3.9375 cubic ft, so yes, the area will greater than 3 and 1/2 ft.

Hope this helps:)

Step-by-step explanation:

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2 years ago

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How many solutions does a triangle with values a=42, A=117 degrees, and b=34 have?

tatyana61 [14]

Answer:one solution

Step-by-step explanation:

Given

a=42

b=34

angle A=117

using sine rule

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

$\frac{42}{sin117}=\frac{34}{sinB}$

$sinB=\frac{34}{42}\times sin(117)$

sinB=0.7212

$B=46.161 ^{\circ}$

and A+B+C=180

$C=16.839^{\circ}$

Thus 2 triangles can be formed with given value

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2 years ago

The number of minutes taken for a chemical reaction if f(t,x). it depends on the temperature t degrees celcius, and the quantity

tatiyna

Hello,

Just for the fun, it's been a long time for that.

$f(30,5)=50\\

f(t+3,x)=f(t,x)-5\ ==\textgreater \dfrac{f(t+3,x)-f(t,x)}{3}= -\frac{5}{3} \\

f(t,x+2)=f(t,x)-3\ ==\textgreater\dfrac{f(t,x+2)-f(t,x)}{2}= -\frac{3}{2} \\

f(t+\Delta t,\Delta x)=f(f,t)+ \dfrac{\partial{f(t,x)}}{\partial{t}} *\Delta t+ \dfrac{\partial{f(t,x)}}{\partial{x}} *\Delta x\\

f(33,4)\approx f(30,5)- \dfrac{5}{3} *3-\dfrac{3}{2}*(-1)= 50- 5+1.5=46.5\\


$

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2 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,

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

Hence, the mat is 4.33 ft high off the ground.

5 0

2 years ago

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