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2 years ago

Select all triangles whose side lengths prove they are right triangles. You must select all that apply.

Mathematics

2 answers:

d1i1m1o1n [39]2 years ago

7 0

Answer:

B and D

Step-by-step explanation:

0.5 squared + 1.2 squared = 1.69 $\sqrt{}$ 1.69 = 1.3

3 squared plus 4 squared = 25 $\sqrt{}$ 25 = 5

if you need help with a question similar to this, you can use Desmos Calculator. hopes this helps :)

Bogdan [553]2 years ago

6 0

It is D.

Reason. 3^2 is 9, 4^2 is 16.
16+9= 25
The square root of 25 is 5.
Therefore it is D.

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The scatter plot shows the wages of employees.

Paladinen [302]

(1) The expected number of copies sold is 75

(2) The expected wage is $720

Explanation:

From the graph, we can see that x represents the number of copiers sold.

y represents the wages.

The equation of the graph is $y=6x+120$

(1) Given that an employee earn $570.

We need to determine the number of copiers sold.

Let us substitute $y=570$ in the equation $y=6x+120$ , we get,

$570=6x+120$

$450=6x$

$75=x$

Thus, the number of copiers sold is 75

(2) Given that if an employee sells 100 copiers, we need to determine the expected wage.

Let us substitute $x=100$ in the equation $y=6x+120$ , we get,

$y=6(100)+120$

$y=600+120$

$y=720$

Thus, the expected wage is $720

7 0

4 years ago

A 160-inch strip of metal 20 inches wide is to be made into a small open trough by bending up two sides on the long side, at rig

babymother [125]

Answer:

The value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.

Step-by-step explanation:

The dimensions of given metal strip are

Length = 160 inch

Width = 20 inch

Let the side bend x inch from each sides to make a open box.

Dimensions of the box are

Length = 160-2x inch

Breadth = 20-2x inch

Height = x inch

The volume of a cuboid is

$V=length\times breadth \times height$

Volume of box is

$V(x)=(160-2x)\times (20-2x)\times x$

$V(x)=(160-2x)(20-2x)x$

$V(x)=4 x^3 - 360 x^2 + 3200 x$

Differentiate with respect to x.

$V'(x)=12x^2 - 720 x + 3200$

Equate V'(x)=0, to find the critical points.

$0=12x^2 - 720 x + 3200$

Using quadratic formula,

$x=30\pm 10\sqrt{\frac{\left(19\right)}{3}}$

The critical values are

$x_1=30+10\sqrt{\frac{\left(19\right)}{3}}\approx 55.166$

$x_2=30-10\sqrt{\frac{\left(19\right)}{3}}\approx 4.834$

Differentiate V'(x) with respect to x.

$V'(x)=24x - 720$

The value of double derivative at critical points are

$V'(55.166)=24(55.166) - 720=603.984$

$V'(4.834)=24(4.834) - 720=-603.984$

Since the value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.

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aleksley [76]

Answer:

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