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2 years ago

Good afternoon, can you help me please

Mathematics

1 answer:

Alexus [3.1K]2 years ago

8 0

It would just be 14-10 each week so 14-10 14-10 14-10 and so on

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Solve system of equation 4b-2d=5, -2b+d=1

Fed [463]

Answer:

no solution

Step-by-step explanation:

3 0

2 years ago

Please answer this correctly which is the correct options

MakcuM [25]

The first two are equivalent to 15 = k + 13

8 = k + 13 - 7 → ( add 7 to both sides )

8 + 7 = k + 13 ⇒ 15 = k + 13 ' is equivalent '

12 = k + 13 - 3 → ( add 3 to both sides )

12 + 3 = k + 13 ' is equivalent '

4 0

2 years ago

When StartFraction 2 x squared Over y EndFraction = StartFraction w + 2 Over 4 EndFraction is solved for w, one equation is w =

Andru [333]

Answer:

the first one

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

seven different gifts are to be distributed among 10 children. How many distinct results are possible if no child is to receive

Alik [6]

Answer:

120 distinct results are possible if no child is to receive more than one gifts.

Step-by-step explanation:

When the order is not important, we use the combination formula:

$C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The order is said to be not important if for example, John receiving the Buffalo Bills jersey and then Laura receiving the Cleveland Browns jersey is the same as Laura receiving the Cleveland Browns jersey before John receives the Buffalo Bills jersey.

In this problem, we have that:

Combinations of 7 from a set of 10 elements. So

$C_{n,x} = \frac{n!}{x!(n-x)!}$

$C_{10,7} = \frac{10!}{7!(3)!} = 120$

120 distinct results are possible if no child is to receive more than one gifts.

6 0

2 years ago

I really am confused on how to write a proof.

kvasek [131]

Since ABCD is a parallelogram, the opposite sides will be parallel and equal,

$\begin{gathered} AB=CD \\ BC=AD \end{gathered}$

Consider that AC acts as a transversal to the parallel lines AB and CD, so we can write,

$\begin{gathered} \angle CAD=\angle ACB\text{ (Alternate Interior Angles)} \\ BC=AD\text{ (Opposite sides of parallelogram)} \\ \angle ADB=\angle CBD\text{ (Alternate Interior Angles)} \end{gathered}$

So by the ASA criteria, the triangle AED is congruent to the triangle CEB,

Then the corresponding parts of the triangles will be equal,

$\begin{gathered} AE=CE \\ BE=DE \end{gathered}$

Hence Proved.

8 0

9 months ago