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3 years ago

Please help, solve a and b

Mathematics

2 answers:

Troyanec [42]3 years ago

8 0

Answer:

The distance across the river d=175ft

Step-by-step explanation:

As◇ ABC& ◇ EDC are similar, BC/CD=AB/ DE

4x/7x=100/d

d=100×7/4=175

Darina [25.2K]3 years ago

7 0

Answer:

AE and BD use the same lines for both triangles so the angles for C are the same. this means that all the angles are congruent between both triangles.

Step-by-step explanation:

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Im giving 100 Points to anyone who could solve this for me please.

myrzilka [38]

Answer:

AB = 27.2

BC = 33.5

AC = 50.4

∠A = 38°

∠ABC = 112°

Step-by-step explanation:

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

$\implies \sf \cos(30^{\circ})=\dfrac{29}{BC}$

$\implies \sf BC=\dfrac{29}{\cos(30^{\circ})}$

$\implies \sf BC=\dfrac{58\sqrt{3}}{3}=33.5\:(nearest\:tenth)$

$\implies \sf \tan(30^{\circ})=\dfrac{BD}{29}$

$\implies \sf BD=29\tan(30^{\circ})$

$\implies \sf BD=\dfrac{29\sqrt{3}}{3}$

$\implies \sf \sin(38^{\circ})=\dfrac{BD}{AB}$

$\implies \sf AB=\dfrac{\dfrac{29\sqrt{3}}{3}}{\sin(38^{\circ})}$

$\implies \sf AB=27.2\:(nearest\:tenth)$

$\implies \sf \tan(38^{\circ})=\dfrac{BD}{AD}$

$\implies \sf AD=\dfrac{\dfrac{29\sqrt{3}}{3}}{\tan(38^{\circ})}$

$\implies \sf AD=21.4\:(nearest\:tenth)$

$\implies \sf AC=AD+DC=21.4+29=50.4$

The interior angles of a triangle sum to 180°

⇒ ∠A + 52° + 90° = 180°

⇒ ∠A = 180° - 90° - 52°

⇒ ∠A = 38°

⇒ ∠ABC + 38° + 30° = 180°

⇒ ∠ABC = 180° - 38° - 30°

⇒ ∠ABC = 112°

**I have checked the measures using a graphing programme - see attached**

8 0

2 years ago

The shaded region R in diagram below is enclosed by y-axis, y = x^2 - 1 and y = 3.

Lostsunrise [7]

Cross sections of the volume are washers or annuli with outer radii x(y) + 1, where

y = x(y) ² - 1 ==> x(y) = √(y + 1)

and inner radii 1. The distance between the outermost edge of each shell to the axis of revolution is then 1 + √(y + 1), and the distance between the innermost edge of R on the y-axis to the axis of revolution is 1.

For each value of y in the interval [-1, 3], the corresponding cross section has an area of

π (1 + √(y + 1))² - π (1)² = π (2√(y + 1) + y + 1)

Then the volume of the solid is the integral of this area over [-1, 3]:

$\displaystyle\int_{-1}^3\pi y\,\mathrm dy = \frac{\pi y^2}2\bigg|_{-1}^3 = \boxed{4\pi}$

$\displaystyle\int_{-1}^3 \pi\left(2\sqrt{y+1}+y+1\right)\,\mathrm dy = \pi\left(\frac43(y+1)^{3/2}+\frac{y^2}2+y\right)\bigg|_{-1}^3 = \boxed{\frac{56\pi}3}$

8 0

3 years ago

Please Help!

Deffense [45]

Answer:

Step-by-step explanation:

7 0

3 years ago

I can’t tell if 3 is a dilation or not!

S_A_V [24]

Answer:

No, its not

Step-by-step explanation:

The points of the image are not moved away from the center of dilation proportionally. I don't think anyway.

3 0

3 years ago

In an isometric transformation, the preimage and image must not __________. A. change size B. rotate C. preserve angle measures

hjlf

The answer is most likely A.

3 0

3 years ago