Ask question

Ask question

All categories

3 years ago

10

Dale and 2 friends share one fourth of a cake.Each peice is the same size.Dale belives each peice of cake has an angle measure i

f 45 degrees.Is dale correct?

1 answer:

Umnica [9.8K]3 years ago

4 0

Answer:

The correct answer is Dale is absolutely wrong.

Step-by-step explanation:

When the cake was complete, it had 360°. When the cake is divided into four halves each half now has 90° each.

Three person including Dale is about to eat the one fourth of the cake, divided among them equally.

Thus each friend gets $\frac{90}{3}$ ° = 30° of cake each.

But Dale claims that each piece of cake has an angle measure if 45°, which is a contradiction as each piece is supposed to to be equal.

Thus Dale is WRONG.

You might be interested in

Find the length of BC. <br> A.) 14<br> B.) 9<br> C.) 2<br> D.) 3

Vera_Pavlovna [14]

Answer:

A) 14

Step-by-step explanation:

BC is congruent to DC in saying this you can plug the equations to each other and solve for y and then put its back into the equation for BC to get the length.

3y+5=5y-1

subtract 3y from both sides --> 5=2y-1

add 1 to both sides --> 6=2y

now get y alone, divide by 2 by both side --> y=3

plug y in back to 5y-1 --> 5x3-1

15-1

BC = 14

7 0

3 years ago

The difference between the two roots of the equation 3x^2+10x+c=0 is 4 2/3 . Find the solutions for the equation.

andrezito [222]

Answer:

Given the equation: $3x^2+10x+c =0$

A quadratic equation is in the form: $ax^2+bx+c = 0$ where a, b ,c are the coefficient and a≠0 then the solution is given by :

$x_{1,2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ......[1]

On comparing with given equation we get;

a =3 , b = 10

then, substitute these in equation [1] to solve for c;

$x_{1,2} = \frac{-10\pm \sqrt{10^2-4\cdot 3 \cdot c}}{2 \cdot 3}$

Simplify:

$x_{1,2} = \frac{-10\pm \sqrt{100- 12c}}{6}$

Also, it is given that the difference of two roots of the given equation is $4\frac{2}{3} = \frac{14}{3}$

i.e,

$x_1 -x_2 = \frac{14}{3}$

Here,

$x_1 = \frac{-10 + \sqrt{100- 12c}}{6}$ , ......[2]

$x_2= \frac{-10 - \sqrt{100- 12c}}{6}$ .....[3]

then;

$\frac{-10 + \sqrt{100- 12c}}{6} - (\frac{-10 + \sqrt{100- 12c}}{6}) = \frac{14}{3}$

simplify:

$\frac{2 \sqrt{100- 12c} }{6} = \frac{14}{3}$

or

$\sqrt{100- 12c} = 14$

Squaring both sides we get;

$100-12c = 196$

Subtract 100 from both sides, we get

$100-12c -100= 196-100$

Simplify:

-12c = -96

Divide both sides by -12 we get;

c = 8

Substitute the value of c in equation [2] and [3]; to solve $x_1 , x_2$

$x_1 = \frac{-10 + \sqrt{100- 12\cdot 8}}{6}$

or

$x_1 = \frac{-10 + \sqrt{100- 96}}{6}$ or

$x_1 = \frac{-10 + \sqrt{4}}{6}$

Simplify:

$x_1 = \frac{-4}{3}$

Now, to solve for $x_2$ ;

$x_2 = \frac{-10 - \sqrt{100- 12\cdot 8}}{6}$

or

$x_2 = \frac{-10 - \sqrt{100- 96}}{6}$ or

$x_2 = \frac{-10 - \sqrt{4}}{6}$

Simplify:

$x_2 = -2$

therefore, the solution for the given equation is: $-\frac{4}{3}$ and -2.

3 0

3 years ago

(IF CORRECT WILL MARK BRAINLIEST!)Which of the following equations has exactly one real solution?

telo118 [61]

A
explanation:
2x+10=-6-30
8x+40=0
x=-5

5 0

3 years ago

Could someone please help me asap? Thanks in advance! :)

Elden [556K]

Answer:

A

Step-by-step explanation:

the standard form of a quadratic equation is

ax² + bx + c = 0 : a ≠ 0

obtain (5 + x)(5 - x) = 7 in this form by expanding the factors

25 - x ² = 7 ( subtract 7 from both sides )

18 - x² = 0 ( multiply through by - 1 )

x² - 18 = 0 ← in standard form

with a = 1, b= 0 and c = - 18 → A

4 0

3 years ago

Solve the linear system... y=-3x+5 and 5x-4y=-3

ivolga24 [154]

Answer:

x = 1 , y = 2

Step-by-step explanation:

Solve for substitution

5 0

2 years ago