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Grace [21]
2 years ago
7

What is the value of the expression 4x−y2y+x when x = 3 and y = 3?

Mathematics
1 answer:
kipiarov [429]2 years ago
3 0

Answer:B

Step-by-step explanation:

(4x - y)/(2y + x) when x = 3 and y = 3 is

(4(3) - 3)/(2(3) + 3) = (12 - 3)/(6 + 3) = 9/9 = 1

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How many points are needed to define a ray?<br> 0<br> 1<br> 2<br> 3
beks73 [17]
You only need 1 point to define a ray
5 0
3 years ago
Read 2 more answers
If a=8 and b=4, evaluate the following expression:2a−3b​
Nata [24]

Answer:

4

Step-by-step explanation:

2(8)-3(4)

16-12

4

7 0
3 years ago
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I don’t really get this one.<br> Thx and have a great day!!
gayaneshka [121]

Answer:

The correct option is C). (9,4)

The coordinates of a point N is (9,4)

Step-by-step explanation:

Theory: If point P(x,y) lies on line segment AB and AP: PB=m:n, then we say P divides line AB internally in ratio of m:n and Point is given by

P=(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})

Given that point, M is lying somewhere between point L and point N.

The coordinates of a point L is (-6,14)

The coordinates of a point M is (-3,12)

Also, LM: MN = 1:4

We can write as,

Let,

Point L(-6,14)=(X1, Y1)

Point M(-3,12)=(x,y)

Point N is (X2, Y2)

m=1 and n=4

M(-3,12)=(\frac{mX2+nX1}{m+n} , \frac{mY2+nY1}{m+n})

M(-3,12)=(\frac{1(X2)+4(-6)}{1+4} , \frac{(Y2)+4(14)}{1+4})

M(-3,12)=(\frac{(X2)-24}{5} , \frac{1(Y2)+56}{5})

(-3)=\frac{(X2)-24}{5}

(-15)=X2-24

X2=9

(12)=\frac{(Y2)+56}{5}

(60)=Y2+56

Y2=4

Thus,

The coordinates of a point N is (9,4)

Result: The correct option is C). (9,4)

3 0
3 years ago
Are 3:5and 1:2 equivalent
german

Answer:

No. They are not equivalent.

Step-by-step explanation:

\frac{3}{5} \neq  \frac{1}{2}

=> \frac{6}{10} \neq  \frac{5}{10}

Therefore, 3/5 and 1/2 are not equivalent.

Hoped this helped.

6 0
2 years ago
Read 2 more answers
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
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