Solve the circled ones pls<br> (2,4,6, and 8)

30g of sugar is used to make 4 cakes. How much sugar is used to make 7 cakes

tankabanditka [31]

Answer:

52.5

Step-by-step explanation:

well first you want to find the ratio and to do that just do 30 / 4 which is

7.5 so for every 1 cake you need 7.5 grams of sugar

if you plug in 7 it is just 7 * 7.5

7 * 7.5 = 52.5

so it takes 52.5 grams of sugar to make 7 cakes

hope this helps please make brainliest

4 0

3 years ago

Read 2 more answers

Pls help this is major, ill mark u brainliest

Lera25 [3.4K]

Answer:

105

Step-by-step explanation:

$A=lw$ $A=$ (.5)πr²

to get the area of the shaded parts you have to subtract the area of the semicircle from the area of the rectangle

$A=(16)(9)$

$A=144$

$A=\frac{(3.14)(5)^2}{2}$ $A=\frac{(3.14)(25)}{2}$ $A=\frac{78.5}{2}$

$A=39.25$

subtract the areas

$144-39.25=104.75$

$104.75$ ≅ $105$

4 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .