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2 years ago

Answer question in image attached :) 20 points ! I think it is 2/4 just not 100% sure :/

Mathematics

2 answers:

lord [1]2 years ago

5 0

Answer:

2/4

Step-by-step explanation:

length of BC = 2

length of BD = 4

therefore, BC : BD = 2 : 4 = 2/4

sergiy2304 [10]2 years ago

3 0

<u>Answer</u>:

$\sf \dfrac{BC}{BD} = \dfrac{2}{4}$

<u>Explanation</u>:

BD distance is 4 units

BC distance is 2 units

so in fractions:

$\sf \dfrac{BC}{BD} = \dfrac{2}{4}$

$\sf \dfrac{BC}{BD} = \dfrac{1}{2}$

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Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

podryga [215]

Answer:

Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

G(X) with the graph of F(x)?

O A. The graph of G(X) is the graph of F(x) shifted 9 units try the left.

B. The graph of G(x) is the graph of F(X) shifted 9 units down.

C. The graph of G(x) is the graph of F(x) shifted 9 units up.

D. The graph of G(X) is the graph of F(x) shifted 9 units to the right.

Step-by-step explanation:

Suppose that G(X) = F(x+ 9). Which statement best compares the graph of

G(X) with the graph of F(x)?

O A. The graph of G(X) is the graph of F(x) shifted 9 units try the left.

B. The graph of G(x) is the graph of F(X) shifted 9 units down.

C. The graph of G(x) is the graph of F(x) shifted 9 units up.

D. The graph of G(X) is the graph of F(x) shifted 9 units to the right.

6 0

2 years ago

A phase modulator produces a maximum phase shift of 45 degrees. The modulation frequency range is 300 to 4000 Hz. What is the ma

sergejj [24]

Answer:

Maximum frequency deviation is 2616.29509 Hz

Maximum frequency deviation=(4000-3000)cos45

Max frequency deviation=2616.29509 Hz

8 0

3 years ago

What is the prime factorization of 210

Kazeer [188]

Answer:

= 2*3 * 5 * 7

Step-by-step explanation:

210 = 21* 10

= 3*7 * 5*2

= 2*3 * 5 * 7

8 0

3 years ago

Read 2 more answers

What’s the answer I am very stuck and not sure

NNADVOKAT [17]

the answer is : 3.814697266

4 0

3 years ago

If y= 3x + 6, what is the minimum value of (x^3)(y)?

Rainbow [258]

Answer:

Given the statement: if y =3x+6.

Find the minimum value of $(x^3)(y)$

Let f(x) = $(x^3)(y)$

Substitute the value of y ;

$f(x)=(x^3)(3x+6)$

Distribute the terms;

$f(x)= 3x^4 + 6x^3$

The derivative value of f(x) with respect to x.

$\frac{df}{dx} =\frac{d}{dx}(3x^4+6x^3)$

Using $\frac{d}{dx}(x^n) = nx^{n-1}$

we have;

$\frac{df}{dx} =(12x^3+18x^2)$

Set $\frac{df}{dx} = 0$

then;

$(12x^3+18x^2) =0$

$6x^2(2x + 3) = 0$

By zero product property;

$6x^2=0$ and 2x + 3 = 0

⇒ x=0 and x = $-\frac{3}{2} = -1.5$

then;

at x = 0

f(0) = 0

and

x = -1.5

$f(-1.5) = 3(-1.5)^4 + 6(-1.5)^3 = 15.1875-20.25 = -5.0625$

Hence the minimum value of $(x^3)(y)$ is, -5.0625

3 0

3 years ago