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3 years ago

Coping strategies can help individuals build

Mathematics

2 answers:

attashe74 [19]3 years ago

5 0

Answer:

1 one

Step-by-step explanation:

noname [10]3 years ago

3 0

Answer: Coping strategies can help individuals build resiliency.

Step-by-step explanation:

Resiliency is defined as the capacity to recover quickly from difficulties. Coping with something means finding ways to face and understand the negative feelings that thing causes you and how to overcome them.

The other three options are actually things you would want to learn how to cope with, thus not correct answers.

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Derek says that the quotient of -2/7 divided by -2/21 is -1/3. Part A: What is the correct quotient? Part B: What mistake did De

Thepotemich [5.8K]

Answer:

The correct quotient is 3.

I think Derek has mistaken to divide $-\frac{2}{21}$ by $\frac{2}{7}$ .

Step-by-step explanation:

We have to check the quotient of $(-\frac{2}{7}) \textrm{ divided by }(-\frac{2}{21})$ .

Part A:

Now, $(-\frac{2}{7}) \textrm{ divided by }(-\frac{2}{21})$ .

= $\frac{-\frac{2}{7} }{-\frac{2}{21} } =3$

So, the correct quotient is 3.

Part B:

Derek says that the quotient is $-\frac{1}{3}$ .

So, I think Derek has mistaken to divide $-\frac{2}{21}$ by $\frac{2}{7}$ . (Answer)

8 0

3 years ago

Serggg [28]

Answer:

value of sides YZ is 36 cm

Step-by-step explanation:

Similar triangles states that the length of the corresponding sides are in proportion.

Given that: ΔABC is similar to ΔXYZ

then;

Corresponding sides are in proportion i.e

$\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}$ .....[1]

As per the statement:

side AB = 6 cm, side BC =18 cm and side XY = 12 cm.

Substitute these in [1] to solve for side YZ;

$\frac{6}{12}= \frac{18}{YZ}$

$\frac{1}{2} = \frac{18}{YZ}$

By cross multiply we have;

$YZ = 36$ cm

Therefore, the value of sides YZ is 36 cm

4 0

3 years ago

Read 2 more answers

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$