Answer:
Yes
Step-by-step explanation:
Since the sides are each congruent then ABCD and PQRS are the exact same shape and size. This means the area inside each figure will be the same amount. The areas are the same.
Answer:
y = 1/3x + 5
Step-by-step explanation:
Since the slope-intercept form is y=mx+b, you need to start by rewriting this in slope-intercept form, which you can write this as 3y = 15 + x. You get this by adding X to both sides.
Then, divide everything by Y, which in this case is 3.
3y Divided by 3 is 1, or Y.
15 Divided by 3 is 5.
x, or 1x, divided by three, is 0.3 forever repeating. However, you rewrite this as a fraction, due to slope-intercept form.
That's how you get the final answer, the final answer being y = 1/3x + 5.
Answer:
<h3>
![\boxed{ \bold{ \huge{ \sf{ \frac{2}{15} }}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%20%5Chuge%7B%20%5Csf%7B%20%5Cfrac%7B2%7D%7B15%7D%20%7D%7D%7D%7D)
</h3>
Step-by-step explanation:
![\sf{ \frac{1}{3} \times \frac{2}{5} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5Cfrac%7B1%7D%7B3%7D%20%20%5Ctimes%20%20%5Cfrac%7B2%7D%7B5%7D%20%7D)
▪️To multiply one fraction by another , multiply the numerators for the numerator , and multiply the denominators for its denominator.
⇒![\sf{ \frac{1 \times 2}{3 \times 5} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5Cfrac%7B1%20%5Ctimes%202%7D%7B3%20%5Ctimes%205%7D%20%7D)
⇒![\sf{ \frac{2}{15} }](https://tex.z-dn.net/?f=%20%5Csf%7B%20%5Cfrac%7B2%7D%7B15%7D%20%7D)
Hope I helped!
Best regards!
Answer:
Jenny uses 7⁄12 of the tank of gas.
Step-by-step explanation:
Answer:
The angle formed between CF and the plane ABCD is approximately 47.14°
Step-by-step explanation:
The given parameters are;
BC = 6.8
DE = 9.3
∠BAC = 52°
We note that the angles formed by the vertex of a cuboid are right triangles, therefore, by trigonometric ratios, we get;
sin∠BAC = BC/(The length of a line drawn from A to C)
∴ The length of the line drawn from A to C = BC/sin∠BAC
The length of the line drawn from A to C = 6.8/sin(52°) ≈ 8.63
∴ AC = 8.63
By trigonometry, we have;
The angle formed between CF and the plane ABCD = Angle ∠ACF
![tan\angle X = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length}](https://tex.z-dn.net/?f=tan%5Cangle%20X%20%3D%20%5Cdfrac%7BOpposite%20%5C%20leg%20%5C%20length%7D%7BAdjacent%5C%20leg%20%5C%20length%7D)
![tan\angle ACF = \dfrac{FA}{AC}](https://tex.z-dn.net/?f=tan%5Cangle%20ACF%20%3D%20%5Cdfrac%7BFA%7D%7BAC%7D)
In a cuboid, FA = BG = CH = DE = 9.3
![\therefore tan\angle ACF = \dfrac{9.3}{8.63}](https://tex.z-dn.net/?f=%5Ctherefore%20tan%5Cangle%20ACF%20%3D%20%5Cdfrac%7B9.3%7D%7B8.63%7D)
![\therefore \angle ACF = arctan \left(\dfrac{9.3}{8.63} \right) \approx 47.14^{\circ}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cangle%20ACF%20%3D%20arctan%20%5Cleft%28%5Cdfrac%7B9.3%7D%7B8.63%7D%20%5Cright%29%20%5Capprox%2047.14%5E%7B%5Ccirc%7D)
The angle formed between CF and the plane ABCD = Angle ∠ACF ≈ 47.14°