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GrogVix [38]
3 years ago
13

Positive numbers are located to the right of zero on the number line. True False

Mathematics
2 answers:
Trava [24]3 years ago
7 0

\text {Hi!}

\text {The Answer to this Problem Is:}

\fbox {True!}

\text {When we look at a number line with positive and negative numbers we see} \text {that positives are to the right and negatives are to the left.}

\text {As you can see the number line on the image that all} \text {the negative numbers are to the left and positive numbers are to the right.}

\text {\underline {Remember}}

\text {Positive to the Right.}

\text {Negative to the Left.}

\text {Best of Luck!}

Wewaii [24]3 years ago
4 0

Hey there!☺

Answer:\boxed{\text{True}}

Explanation:

It is true because positive numbers are located to the right of zero. Below zero will make the number negative and it will be located to left. If any number is positive, then it will be located near zero on a number line.

So that means your answer is True.

Hope this helps!☺

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What is partial derivative of z=(2x+3y)^10 with respect to x,y?
maw [93]
Not sure if you mean to ask for the first order partial derivatives, one wrt x and the other wrt y, or the second order partial derivative, first wrt x then wrt y. I'll assume the former.

\dfrac\partial{\partial x}(2x+3y)^{10}=10(2x+3y)^9\times2=20(2x+3y)^9

\dfrac\partial{\partial y}(2x+3y)^{10}=10(2x+3y)^9\times3=30(2x+3y)^9

Or, if you actually did want the second order derivative,

\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8

and in case you meant the other way around, no need to compute that, as z_{xy}=z_{yx} by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because z is a polynomial).
3 0
3 years ago
Read 2 more answers
Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki
Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for f(x) = 3 - \dfrac{1}{2}x from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x

where:

\Delta x = \dfrac{b-a}{a}

a = 2

b =14

n = 6

∴

\Delta x = \dfrac{14-2}{6}

\Delta x = \dfrac{12}{6}

\Delta x =2

Hence;

x_0 = 2 \\ \\  x_1 = 2+2 =4\\ \\  x_2 = 2 + 2(2) \\ \\  x_i = 2 + 2i

Here, we are  using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix}  2+2 (i-1)  \end {pmatrix} \end {pmatrix}2

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2i-2)

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -   \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - 2  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} i

Estimating the integrals, we have :

= 6n - 2 ( \dfrac{n(n-1)}{2})

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0
3 years ago
Is this answer right
ehidna [41]

Answer: No the real answer is 35.10585

Step-by-step explanation:

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svetlana [45]

Answer:i dont know

Step-by-step explanation:

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3 years ago
What is the format of this proof?
elena55 [62]
Divide b from both sides then it should be a equal C therefore if you add be it be the midpoint of a and C because BNB or equal to a C
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