All categories

3 years ago

Find the area of this isosceles triangle.

Mathematics

1 answer:

Drupady [299]3 years ago

7 0

Answer:

use the formula A = 1/2ab(sin C)

Step-by-step explanation:

You might be interested in

Given: △ACE, BD∥AE<br><br> Prove: BA/CB=DE/CD<br><br> Drag

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

Angle 4 = angle 1 and angle 3 =angle 2 Corresponding angles for two parallel lines BD and AE

Triangle AED ||| BDE Angle angle congruent

... By compendo and dividendo

Thus proved

Since two lines are parallel we have corresponding angles equal and also

sides will be proportional in similar triangles

Using that and adjusting compendo and dividendo we get the desired result

6 0

3 years ago

Read 2 more answers

5y = 3478 - 3c if the equation above, c is a constant. If y = 8 is a solution to the equation, what is the value of c?

docker41 [41]

What you would do is simply plug in 8 where y is and solve.
5(8)=3478-3c -Times 5 by 8
40=3478-3c -Add 3c to the opposite side of the equal sign
40+3c=3478 -Subtract 40 from the opposite side of the equal sign
3c=3438 -Divide equation by 3
c=1146

Hope this helps! :)

5 0

3 years ago

(08.07 HC)

andreev551 [17]

Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

$\implies ac=2 \cdot -10=-20$

$\implies b=-1$

Therefore, the two numbers are -5 and 4.

Rewrite $b$ as the sum of these two numbers:

$\implies f(x)=2x^2-5x+4x-10$

Factor the first two terms and the last two terms separately:

$\implies f(x)=x(2x-5)+2(2x-5)$

Factor out the common term (2x - 5):

$\implies f(x)=(x+2)(2x-5)$

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

Therefore:

$\implies (x+2)=0 \implies x=-2$

$\implies (2x-5)=0 \implies x=\dfrac{5}{2}$

So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

The x-value of the vertex is:

$\implies x=\dfrac{-b}{2a}$

Therefore, the x-value of the vertex of the given function is:

$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the <u>leading coefficient</u> of the function is positive, the parabola will open upwards. This is confirmed as the vertex is a minimum point.

The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).