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3 years ago

Select ALL the correct answers.

Mathematics

2 answers:

almond37 [142]3 years ago

7 0

I believe it’s 1, 3 and 4

Morgarella [4.7K]3 years ago

5 0

Answer: only the last two statements are true. AB and CD are parallel. Angles 1 and 3 are corresponding.

Angles 1 &2 3&4 are supplementary. Not corresponding. Check definitions; don't get confused.

Step-by-step explanation:

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The temperature when Spencer arrived at school was a very chilly

nekit [7.7K]

9 degrees. -4 + 13= 9.

6 0

1 year ago

Find

Sedaia [141]

Answer:

The amount of steel used = 94.38 $m^{2}$

Step-by-step explanation:

Height of the tank = 4.5 m.

Diameter of the tank = 4.2 m.

But,

radius = $\frac{diameter}{2}$

= $\frac{4.2}{2}$

radius = 2.1 m

Total surface area of a cylinder = 2 $\pi$ rh + 2 $\pi$ $r^{2}$

= 2 $\pi$ r(r + h)

= 2 x $\frac{22}{7}$ x 2.1 (2.1 + 4.5)

= 44 x 0.3 (6.6)

= 87.12

Total surface area of a cylinder = 87.12 $m^{2}$

Since $\frac{1}{12}$ of the steel was wasted, then;

$\frac{1}{12}$ x 87.12 = 7.26 $m^{2}$

Actual area of steel sheet used = 87.12 + 7.26

= 94.38 $m^{2}$

Thus, 94.38 $m^{2}$ of steel was used in making the tank.

4 0

2 years ago

If h(x) is the inverse of f(x), what is the value of h(f(x))?

Viktor [21]

The value of h(f(x)) would just be x. That is the line y = x which is the line that functions are reflected in to find the inverse.

4 0

2 years ago

Read 2 more answers

The function f(x) = −16x2 + 112 models the height, in feet, of an

natima [27]

For this, just plug in 1.5 into the x variable of this function and solve as such:

$f(1.5)=-16*1.5^2+112\\f(1.5)=-16*2.25+112\\f(1.5)=-36+112\\f(1.5)=76$

<u>The height of the object was at 76 ft.</u>

3 0

2 years ago

Solve the rational equation x divided by 6 equals x squared divided by quantity x minus 1 end quantity, and check for extraneous

Dmitriy789 [7]

Answer:

Solution: $x=-\frac{1}{5}$ , $x=0$

Step-by-step explanation:

Given:

The rational equation to solve is given as:

$\frac{x}{6}=\frac{x^2}{x-1}$

Doing cross product, we get:

$x\times (x-1)=6\times x^2\\x^2-x=6x^2\\\textrm{Bringing all variables to the right side, we get:}\\6x^2-x^2+x=0\\5x^2+x=0\\x(5x+1)=0\\x=0\ or\ 5x+1=0\\x=0\ or\ x=-\frac{1}{5}$

Now, for $x=0$ , the rational equation is equal to 0. So, $x=0$ is a solution.

Also, $x=-\frac{1}{5}$ is also a solution.

So, no extraneous solution.

4 0

3 years ago