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lorasvet [3.4K]

3 years ago

6

a student bought a juice pouch for $2.50 and 3 bags of chips. the total cost was $5.05 write and solve an equation to determine

the cost of a bag of chips. please show your work !!! thank you.

2 answers:

slavikrds [6]3 years ago

7 0

Answer:

$0.85

Step-by-step explanation:

Let x = cost of 1 bag of chips

3x + 2.50 = 5.05

3x = 2.55

x = 0.85

I hope this helped and please mark me as brainliest!

Wittaler [7]3 years ago

5 0

First, write it as an equation

2.50+3x=5.05

Next, subtract 2.50 from both sides

2.50-2.50+3x=5.05-2.50

3x=2.55

Divide 3 on both sides

3x/3=2.55/3

x=0.85

The cost of a bag of chips is 0.85

hope i helped :D

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3 years ago

Principal = 5300=5300 rupees

PSYCHO15rus [73]

Answer:

Using the formula:

$A=P+I$

where

A is the total amount

P is the principal

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As per the statement:

Principal(P) = 5300 rupees.

rate of interest(r) = 6.5% = 0.065

Total amount(A) = 6678 rupees.

Then using above formula we have;

$6678 = 5300+I$

Subtract 5300 both sides we get;

$1378 = I$

or

$I = \$1378$

We have to find the time period.

Using formula of Simple interest:

$I = Prt$

where r is the rate of interest (in decimal)

here, r = 6.5% = 0.065

Substitute the given values top find t:

$1378 = 5300 \cdot 0.065 \cdot t$

⇒ $1378 = 344.5t$

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4 years ago

L :V --> W is a linear transformation. Prove each of the following (a) ker L is a subspace of V. (b) range L is a subspace of

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Answer:

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<em>Second.</em> Let us prove that $\alpha x\in\ker L$ . By linearity

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And the statement readily follows.

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<em>First.</em> Let us prove that range of $L$ is not empty. This is easy because $L(0_V)=0_W$ , by linearity.

<em>Second.</em> Let us prove that $\alpha u$ is on the range of $L$ .

$\alpha u = \alpha L(x) = L(\alpha x) = L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)=\alpha u$ . Thus $\alpha u$ is in the range of $L$ .

<em>Third.</em> Let us prove that $u+v$ is in the range of $L$ .

$u+v = L(x)+L(y) = L(x+y)=L(z)$ .

Then, there exist an element $z\in V$ such that $L(z)= u +v$ . Thus $u +v$ is in the range of $L$ .

Notice that in this second part of the problem we used the linearity in the reverse order, compared with the first part of the exercise.

Step-by-step explanation:

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Ivanshal [37]

Answer:

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