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3 years ago

11

Can someone help me w answer 2 & 5. And can u check my other answers if there correct ? Would appreciate it

1 answer:

slega [8]3 years ago

5 0

2.A third of a cubed
5. Three b squared plus two a cubed
And all your other answers are correct

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Evan goes out to dinner with some friends to celebrate their graduation. Their total bill is $34.08. If the 3 of them split the

Novosadov [1.4K]

Answer:

each will pay 11.36

Step-by-step explanation:

34.08 divided by 3 is 11.36

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2 years ago

Anne and Kyleigh worked over summer vacation.

juin [17]

Kyleigh earned more money than Anne

7 0

3 years ago

In math class the girl to boy ratio is 8 to 6 if there are 24 girls in the class , how many boys are there ?

Volgvan

Answer: The required number of boys in the class is 18.

Step-by-step explanation: Given that in math class, the girl to boy ratio is 8 to 6 and there are 24 girls in the class.

We are to find the number of boys in the class.

Let 8x and 6x represents the number of girls and boys in the class.

Then, according to the given information, we have

$8x=24\\\\\Rightarrow x=\dfrac{24}{8}\\\\\Rightarrow x=3.$

Therefore, the number of boys in the class is given by

$6x=6\times3=18.$

Thus, the required number of boys in the class is 18.

7 0

3 years ago

The vertex of a parabola is at (6,5), and its axis of symmetry is vertical. One of the x-intercepts is at (8, 0). What is the x-

Elena-2011 [213]

Answer:

$x=4$

Step-by-step explanation:

Vertex form of the quadratic equation: $f(x) = a(x - h)^2+k$

where the vertex is $(h, k)$

Given vertex = (6, 5):

$\implies f(x) = a(x - 6)^2+5$

If one the x-intercepts is (8, 0):

$\implies f(8) =0$

$\implies a(8 - 6)^2+5=0$

$\implies 4a+5=0$

$\implies a=-\dfrac{5}{4}$

Therefore, equation of parabola:

$\implies f(x) = -\dfrac54(x - 6)^2+5$

x-intercepts occur when f(x) = 0:

$\implies f(x)=0$

$\implies -\dfrac54(x - 6)^2+5=0$

$\implies -\dfrac54(x - 6)^2=-5$

$\implies (x - 6)^2=4$

$\implies x-6=\pm 2$

$\implies x=8, x=4$

Therefore, the x-coordinate of the other x-intercept is 4.

5 0

2 years ago

Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

<u>Area</u>

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

<u>Perimeter</u>

Arc length = $r\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

<u>Shape ABCD</u>

<u>Area</u>

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

<u>Perimeter</u>

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago