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Vesnalui [34]
2 years ago
14

The graph below shows Sean's distance, d, from his home (in kilometers) as a function of the time, t (in hours).

Mathematics
1 answer:
sdas [7]2 years ago
4 0

Answer:

20

20

5

Step-by-step explanation:

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AB, CD, and EF intersect at O. If CD is perpendicular to EF and m
REY [17]

Answer:

m(∠AOF) = 148°

Step-by-step explanation:

From the figure attached,

CD intersects line EF at a point O.

Line CD is perpendicular to the line EF.

m(∠AOE) = 32°

m(∠COE) = 90°

Since m(∠COE) = m(∠AOE) + m(∠AOC) = 90°

32° + m(∠AOC) = 90°

m(∠AOC) = 90° - 32° = 58°

m(∠AOF) = m(∠AOC) + m(∠COF)

               = 58° + 90°

               = 148°

Therefore, m(∠AOF) = 148° will be the answer.

4 0
3 years ago
What is the side lengths, in inches, of a cube with a volume of 1 1 cubic inch? Branliest
Mnenie [13.5K]

Answer:

34

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the coordinates of the vertices of each figure after the given transformation.
zhannawk [14.2K]

Answer:

uhm sorry this is really hard maybe answer 2

Step-by-step explanation:

7 0
3 years ago
Use Euclid's division algorithm to find the HCF of 441, 567, 693
Illusion [34]
Let a = 693, b = 567 and c = 441

Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under

693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63

Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63

441 = 63 x 7+0
=> HCF of 441 and 63 is 63.

Hence, HCF of 441, 567 and 693 is 63.
6 0
3 years ago
Apply Newton’s method to estimate the solution of x3 − x − 1 = 0 by taking x1 = 1 and finding the least n such that xn and xn +
ss7ja [257]

Solution :

Given

$f(x)=x^3-x-1, x_1=1$

$f'(x)=3x^2-1$

Let the initial approximation is $x_1 =1$

So by Newton's method, we get

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$

$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$

$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$

$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$

$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$

$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$

$x_5 \approx x_6$ are identical up to eight decimal places.

The approximate real root is x ≈ 1.32471795

∴ x = 1.32471795

4 0
3 years ago
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