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Sergeu [11.5K]
3 years ago
10

An engineering scale model shows a church that is 2 inches tall. if the scale is 1 inch = 265 feet, how tall is the actual churc

h?
Mathematics
1 answer:
MAXImum [283]3 years ago
7 0
Well if one inch = 265 feet then two inches = 530 feet. So the Actual church is 530 ft. 
Hope this helps! And if you ever need anything else just ask :)
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For which equations is 8 a solution? Select the four correct answer
maxonik [38]

Answer:

I <em>believe</em> the answers are= 2, 3, 6, 8

Step-by-step explanation:

For #2, you would subtract 2 from both sides. 10-2=8. x=8.

For #3, you would add 4 to both sides of the equation, making the final answer: x=8

For #6, you would divide both sides by 3. 24 divided by three is 8.

For #8, I multiplied the whole equation by 8, which is the denominator of x/8. 1 times 8=8. The final answer would be x=8.

Hope this helps!  

8 0
2 years ago
Read 2 more answers
What inequality represents the verbal expression? all real numbers less than 70
neonofarm [45]

For this case we have:

Let x be the variable that belongs to the real numbers. Then, all reals less than 70 can be expressed as:

x

The tip of the inequality is directed to the real numbers, since they tell us that they are less than 70, for 70 the inequality remains open.

Answer:

x

5 0
3 years ago
249798.828806 rounded to the nearest ten
NNADVOKAT [17]

Answer: 249800

Step-by-step explanation:

Uhh you round it.

5 0
2 years ago
EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =
geniusboy [140]

The Lagrangian,

L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)

has critical points where its partial derivatives vanish:

L_x=1-\lambda-2\mu x=0

L_y=2+\lambda-2\mu y=0

L_z=9-\lambda=0

L_\lambda=x-y+z-1=0

L_\mu=x^2+y^2-1=0

L_z=0 tells us \lambda=9, so that

L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu

L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}

Then with L_\mu=0, we get

x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2

and L_\lambda=0 tells us

x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}

Then there are two critical points, \left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right). The critical point with the negative x-coordinates gives the maximum value, 9+\sqrt{185}.

8 0
3 years ago
robert plant 7 trees behind westwood school. he planted 6 times as many trees in front of the school. how many trees did he plan
Sati [7]
49 in total or 42 trees in the front

4 0
3 years ago
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