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mario62 [17]
3 years ago
5

Find the midpoint.........

Mathematics
1 answer:
patriot [66]3 years ago
5 0

Answer:

(14,2)

Step-by-step explanation:

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Corgurent or supplementary
nataly862011 [7]

Answer:

Congruent:

B, E, F, G

The rest are supplementary.

Step-by-step explanation:

Mark me Brainliest?

5 0
3 years ago
A flagpole casts a shadow of 10ft long. A girl standing nnext to the flagpole casts a shadow 2.5 ft. long.If the girl is 5 ft. t
loris [4]
Since we know that they have to e similar, it is a 20ft tall flagpole. the shadow of the girl is half of her height, so since the flag pole shadow is 10ft it is simple 10 multiplied by 2
3 0
3 years ago
Read 2 more answers
What number should be placed in the box to help complete the division calculation? can someone please do the problem (8587/12) s
disa [49]

Answer:

8587 should be in the box

Step-by-step explanation:

we would see it question as:       12/- 8587

12 goes into 85  7 times, so 7

12 goes into 18 1 time, so 1

12 goes into 67 5 times, so 5

then the remainder is 7

or you further divide to get . 5 8 3333333...

so your answer should be 715r7 or 715.58333

5 0
3 years ago
PLEASE HELP!! brainliest if correct!!
Sonja [21]
Graph #1: No
Graph #2: Yes
Graph #3: No
Graph #4: No
Graph #5: No
Graph #6: Yes

Reasoning:

The vertical line test is a test that determines wether a graph is a function or a relation. The vertical line test shows that if you construct a vertical line through any point on the graph, then the vertical line should only intercept the graph once for it to be a function.
6 0
2 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
3 years ago
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