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faltersainse [42]

2 years ago

13

Use the proportion of the triangle enlargement to find the missing measure of the enlarged triangle. 1:Set up the proportion: 9/

6 = x/16 2: use cross product: 9(16)=6x 3: Simplify: 144= ____4: Divide: ___=x

2 answers:

Archy [21]2 years ago

8 0

Answer:

<h2>x = 24</h2>

Step-by-step explanation:

The given proportion is

$\frac{9}{6}=\frac{x}{16}$

Then, we solve for $x$ . First, we need to use cross product

$\frac{9}{6}=\frac{x}{16}\\9(16)=x(6)\\144=6x$

Then, we isolate $x$ and simplify

$6x=144\\x=\frac{144}{6}\\ x=24$

Therefore, the answer to this proportion is 24.

ZanzabumX [31]2 years ago

5 0

The value of x = 24

Step-by-step explanation:

The proportion is set as ;

$\frac{9}{6} =\frac{x}{16}$

Applying cross product

9*16=6*x

144=6x --------simplify by dividing by 6 both sides

144/6 = 6x/6

24=x

Learn More

Proportion in Enlargement : brainly.com/question/9930004

Keywords : proportion, enlargement ,triangles, cross products

#LearnwithBrainly

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Now imagine that instead of walking along the path 1→2→3→4→1, ann walks 80 meters on a straight line 33∘ north of east starting

stealth61 [152]

Answer:

Anna's walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$ and refer attachment.

Step-by-step explanation:

Let the origin be the point 1 from where Ann start walking.

Ann walks 80 meters on a straight line 33° north of the east starting at point 1 as shown in figure below,

Resolving into the vectors, the vertical component will be 80Sin33° and Horizontal component will be 80Cos33° as shown in figure (2)

Ann walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$

Thus, Anna's walk as a vector representation is $80\cos 33^{\circ}\hat{i}+80 \sin33^{\circ}\hat{j}$

7 0

2 years ago

Read 2 more answers

What is the factored form of the binomial expansion 125x^3 + 525x^2 + 735x + 343?

yarga [219]

Answer:

Step-by-step explanation:

1. regroup terms

$125x^3+343+525x^2+735x$

2. Rewrite $125x^3$ as $(5x)^3$ and 343 as $7^3$

$(5x)^3+7^3+525x^2+735x$

3. Since both terms are perfect cubes, factor using the sum of cubes formula, $a^3+b^3 = (a+b)(a^2-ab+b^2)$ , where $a=5x$ and $b=7$

$(5x+7)((5x)^2-(7)(5x)+(7)^2)+525x^2+735x$

$(5x+7)(25x^2-35x+49)+525x^2+735x$

4. Factor $105x$ out of $525x^2+735x$

$(5x+7)(25x^2-35x+49)+105x(5x+7)$

5. Regroup

$(5x+7)(25x^2-35x+49+105x)$

$(5x+7)(25x^2+70x+49)$

6. Factor again

$(5x+7)(5x+7)^2$

$(5x+7)^3$

4 0

2 years ago

Read 2 more answers

Write an expression involving integers for each statement a) moving 4 steps left, then moving 9 steps right b) on 3 separate occ

Charra [1.4K]

Answer:

a) x-4+9

b) x-2

For part b, I am not 100% sure about my answer, but I am sure about part a.

3 0

2 years ago

A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

Scrat [10]

Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

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