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3 years ago

Solving multi-step equations

Mathematics

1 answer:

astra-53 [7]3 years ago

4 0

Please elaborate so I can help you. For example, give me an example problem

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Solve x+6(x-1)=7(3+x)

Lina20 [59]

Let's solve your equation step-by-step.

x+6(x−1)=7(3+x)

Step 1: Simplify both sides of the equation.

x+6(x−1)=7(3+x)

x+(6)(x)+(6)(−1)=(7)(3)+(7)(x)(Distribute)

x+6x+−6=21+7x

(x+6x)+(−6)=7x+21(Combine Like Terms)

7x+−6=7x+21

7x−6=7x+21

Step 2: Subtract 7x from both sides.

7x−6−7x=7x+21−7x

−6=21

Step 3: Add 6 to both sides.

−6+6=21+6

0=27

5 0

3 years ago

WILL GIVE BRAINLIEST, AMAZING RATING AND THANKS!

natita [175]

it goes up 0.8 every time

6 0

3 years ago

Read 2 more answers

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$

$KA(\frac{h}{x}-1)=AP$

$ND(\frac{h}{x}-1)=PD$

Adding above two, we get

$(KA+ND)(\frac{h}{x}-1)=(AP+PD)$

⇒ $(KN-AD)=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x^2}{h-x}$

⇒ $x^2=ah-ax-xh+x^2$

⇒ $x(h+a)=ah$

⇒ $x=\frac{ah}{a+h}$

3 0

3 years ago

Find the value of the trigonometric ratio. A.3/5 B.3/4 C.4/3 D.4/5

Montano1993 [528]

We know that :

$tan(X) = \dfrac{ZY}{XY}$

$tan(X) = \dfrac{15}{20}$

$tan(X) = \dfrac{3}{4}$

therefore , the correct option is B. 3/4

7 0

3 years ago

Read 2 more answers

62% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

dedylja [7]

Answer:

a) 0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b) 0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c) 0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d) 0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

Step-by-step explanation:

To solve this question, we use the binomial probability distribution, and also it's approximation to the normal distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

62% of owned dogs in the United States are spayed or neutered.

This means that $p = 0.62$

48 owned dogs are randomly selected

This means that $n = 48$

Mean and standard deviation, for the approximation:

$\mu = E(x) = np = 48*0.62 = 29.76$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{48*0.62*0.38} = 3.36$

a. Exactly 30 of them are spayed or neutered.

This is P(X = 30), which is not necessary the use of the approximation.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 30) = C_{48,30}.(0.62)^{30}.(0.38)^{18} = 0.1180$

0.1180 = 11.80% probability that exactly 30 of them are spayed or neutered.

b. At most 33 of them are spayed or neutered.

Now we use the approximation. This is, using continuity correction, $P(X \leq 33 + 0.5) = P(X \leq 33.5)$ , which is the pvalue of Z when X = 33.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33.5 - 29.76}{3.36}$

$Z = 1.11$

$Z = 1.11$ has a pvalue of 0.8665

0.8665 = 86.65% probability that at most 33 of them are spayed or neutered.

c. At least 31 of them are spayed or neutered.

Using continuity correction, this is $P(X \geq 31 - 0.5) = P(X \geq 30.5)$ , which is 1 subtracted by the pvalue of Z when X = 30.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

1 - 0.5871 = 0.4129

0.4129 = 41.29% probability that at least 31 of them are spayed or neutered.

d. Between 24 and 30 (including 24 and 30) of them are spayed or neutered.

This is, using continuity correction, $P(24 - 0.5 \leq X \leq 30 + 0.5) = P(23.5 \leq X \leq 30.5)$ , which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 23.5.

X = 30.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{30.5 - 29.76}{3.36}$

$Z = 0.22$

$Z = 0.22$ has a pvalue of 0.5871

X = 23.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{23.5 - 29.76}{3.36}$

$Z = -1.86$

$Z = -1.86$ has a pvalue of 0.0314

0.5871 - 0.0314 = 0.5557

0.5557 = 55.57% probability that between 24 and 30 of them are spayed or neutered.

8 0

3 years ago