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3 years ago

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On Saturday Keisha ran 3.218 kilometers. on Sunday she ran 2.41 kilometers. How much further did she run on Saturday than on Sun

day

2 answers:

Len [333]3 years ago

7 0

0.808 km, you find the difference between the two numbers

worty [1.4K]3 years ago

7 0

Answer:

0.808 kilometers

Step-by-step explanation:

Given: On Saturday Keisha ran $3.218$ $\text{kilometer}$ . on Sunday she ran $2.41$ $\text{kilometer}$ .

To Find: How much further did she run on Saturday than on Sunday.

Solution:

Kilometers run by keisha on saturday $=3.218$ $\text{kilometer}$

Kilometers run by keisha on sunday $=2.41$ $\text{kilometer}$

Now,

kilometers covered further by keisha on saturday than sunday

$=\text{kilometers covered on Saturday}-\text{kilometers covered on Sunday}$

putting values,

$3.218-2.41$

$0.808$ $\text{kilometer}$

keisha run $0.808$ $\text{kilometer}$ further on saturday than on sunday.

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If 50% of a number is 18 what is 75% of the number

den301095 [7]

Answer:

A) 50% of 36 is 18

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Step-by-step explanation:

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3 years ago

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Test the set of polynomials for linear independence. If it is linearly dependent, express one of the polynomials as a linear com

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Answer:

The set of polynomial is Linearly Independent.

Step-by-step explanation:

Given - {f(x) =7 + x, g(x) = 7 +x^2, h(x)=7 - x + x^2} in P^2

To find - Test the set of polynomials for linear independence.

Definition used -

A set of n vectors of length n is linearly independent if the matrix with these vectors as columns has a non-zero determinant.

The set is dependent if the determinant is zero.

Solution -

Given that,

f(x) =7 + x,

g(x) = 7 +x^2,

h(x)=7 - x + x^2

Now,

We can also write them as

f(x) = 7 + 1.x + 0.x²

g(x) = 7 + 0.x + 1.x²

h(x) = 7 - 1.x + 1.x²

Now,

The coefficient matrix becomes

A = $\left[\begin{array}{ccc}7&1&0\\7&0&1\\7&-1&1\end{array}\right]$

Now,

Det(A) = 7(0 + 1) - 1(7 - 7) + 0

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3 years ago

☆pls help no one has helped me and I'm so confused!

motikmotik

Answer:

Step-by-step explanation:

a). $\frac{5\pm\sqrt{-4}}{3}$ = $\frac{5}{3}\pm \frac{\sqrt{-4}}{3}$

= $\frac{5}{3}\pm\frac{2\sqrt{(-1)} }{3}$

= $\frac{5}{3}\pm\frac{2i}{3}$ [Since, i = $\sqrt{(-1)}$ ]

b). $\frac{10\pm\sqrt{-16}}{2}$ = $\frac{10}{2}\pm \frac{\sqrt{-16}}{2}$

= $5\pm \frac{4\sqrt{-1}}{2}$

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c). $\frac{-3\pm \sqrt{-144}}{6}$ = $-\frac{3}{6}\pm \frac{\sqrt{-144}}{6}$

= $-\frac{3}{6}\pm \frac{12\sqrt{-1}}{6}$

= $-\frac{1}{2}\pm 2\sqrt{-1}$

= $-\frac{1}{2}\pm 2i$ [Since, i = $\sqrt{(-1)}$ ]

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3 years ago

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Sergio039 [100]

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2 years ago

WORK OUT (8*10^-3)*(2*10^-4)

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2 years ago

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