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storchak [24]
2 years ago
14

4y+2x=180 solve for x and y

Mathematics
1 answer:
kirill [66]2 years ago
5 0
There you go!! Hope it helps

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Help pleaseeeeeee quickly
shusha [124]

The first sentence in the chart belongs to the subjunctive mood, while the other sentences can be classified as part of the conditional mood.

<h3>What is a verb mood?</h3>

In language, the verb mood indicates the attitude of the speaker, and therefore, it provides clues about the way language is used. For example, the imperative mode indicates an order is being given.

<h3>What is the difference between the conditional and the subjunctive mood?</h3>
  • Subjunctive: It is used for expressing wishes; due to this, it is common to find the use of words such as "wish".
  • Conditional: It expresses a condition or a situation that will/can happen if another situation occurs. This can be identified due to the use of "if".

Based on this, the first sentences belongs to the subjunctive mood, while the other sentences are part of the conditional.

Learn more about the subjunctive in: brainly.com/question/22728240

#SPJ1

6 0
2 years ago
74% of 643 is what number
oksano4ka [1.4K]
475.82
is the answr i think

7 0
2 years ago
Read 2 more answers
Simplify the expression.
Maksim231197 [3]

Answer:

6.8x - 5

Step-by-step explanation:

To answer this question I combined like terms:

(2.5x + 4.3x) - 5

2.5x + 4.3x = 6.8x

Then you add the 5 to the end of the expression:

6.8x - 5

7 0
2 years ago
Estimate the quotient to the nearest whole number by using compatible numbers 32.44 ÷ 4.36
GarryVolchara [31]
I'm pretty sure its 7
7 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
2 years ago
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