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3 years ago

How to find a slope

Mathematics

1 answer:

exis [7]3 years ago

5 0

Answer:

Step-by-step explanation:

y₂₋y₁ / x₂-x₁ if you have two points you can plug them in and find the slope using this.

If it is on a graph you can look at two points and count the rise over run, for example if one point is (0, 0) and another is (2, 3) then from the first point it goes up 2 and over 3, so the slope would be 2/3

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A right triangle has a hypotenuse of 70 feet and a leg of 35 feet. What is the length of the other leg? Select one:

saveliy_v [14]

H = 70ft.
P = 35ft.
B = ?

H^2 = P^2 + B^2
70×70 = (35×35) + B^2
B^2 = 4900-1225
B = √3675
B = 35√3 ft.

3 0

3 years ago

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

melamori03 [73]

Answer:

$6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}$

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs $AB=3$ , $BC=4$ , and $AC=5$ . The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base $b$ and height $h$ is given by $A=\frac{1}{2}bh$ . Therefore, the area of this right triangle is:

$A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}$

The other triangle is a bit trickier. Triangle $\triangle ADC$ is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

$A=\sqrt{s(s-a)(s-b)(s-c)}$ , where $a$ , $b$ , and $c$ are three sides of the triangle and $s$ is the semi-perimeter ( $s=\frac{a+b+c}{2}$ ).

The semi-perimeter, $s$ , is:

$s=\frac{5+5+4}{2}=\frac{14}{2}=7$

Therefore, the area of the isosceles triangle is:

$A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}$

Thus, the area of the quadrilateral is:

$6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}$

4 0

3 years ago

When BA=10 ft, find the area of the region that is NOT shaded. Round to the nearest whole number. (reply so i can draw it out, p

nexus9112 [7]

The area of the region that is not shaded in the diagram is determined through the following solution:

A = (1 − (60°/360°)*π) * (10 ft ) ^2 A = 56*π*100 [ft2] A = 250/3 π [ft2]
A = 262 [ft2]

5 0

3 years ago

Simplify the problem: Sin theta Sec Theta

kherson [118]

Answer:

$\tan \theta$