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3 years ago

Help me I need help this is stressful

Mathematics

1 answer:

Alisiya [41]3 years ago

5 0

Answer:

"You could add each number one by one in your head, and the total is your answer."

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-4x-3(-44-23)=13 please show the work, i’ll mark brainliest

pychu [463]

Answer:

Step-by-step explanation:

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

−4x−3(−44−23)=13

Step 1: Simplify both sides of the equation.

−4x+201=13

Step 2: Subtract 201 from both sides.

−4x+201−201=13−201

−4x=−188

Step 3: Divide both sides by -4.

−4x

−4

−188

−4

x=47

3 0

2 years ago

Evaluate: 6-(2/3)^2 A. 17/3 B. 52/3 C. 49/9 D. 50/9

Komok [63]

Answer:

The correct answer is: "Option [D]".

Step-by-step explanation:

Hi student, let me help you out!

....................................................................................................................................

Let's use the acronym PEMDAS. With the help of this little acronym, we will not make mistakes in the Order of Operations! :)

$\dag\textsf{Acronym \: PEMDAS}$

P=Parentheses,

E=Exponents,

M=Multiplication,

D=Division,

A=Addition,

S=Subtraction.

Now let's start evaluating our expression, which is $\mathsf{6-(\cfrac{2}{3})^2}$

According to PEMDAS, the operation that we should perform is "E-Exponents".

Notice that we have a fraction raised to a power. When this happens, we raise both the numerator (2 in this case) and the denominator (3 in this case) to that power, which is 2. After this we obtain $\mathsf{6-\cfrac{4}{9}}$ .

See, we raised both the numerator and the denominator to the power of 2.

Now what we should do is subtract fractions.

Note that 6 and -4/9 have unlike denominators. First, let's write 6 as a fraction: $\mathrm{\cfrac{6}{1}-\cfrac{4}{9}}$ . Now let's multiply the denominator and the numerator of the first fraction times 9: $\mathrm{\cfrac{54}{9}-\cfrac{4}{9}}$ .

See, now the fractions have the same denominator. All we should do now is subtract the numerators: $\mathrm{\cfrac{50}{9}}$ .

∴, the answer is Option D.

Hope this helped you out, ask in comments if any queries arise.

Best Wishes!

$\star\bigstar\underline{\overline{\overline{\underline{\textsf{Reach \: far. Aim \: high. Dream \: big.}}}}}\bigstar\star$

$\underline{\rule{300}{5}}$

5 0

2 years ago

Read 2 more answers

Alan bought two bikes. He sold one to Beth for $300 taking a 25% loss. He also sold one to Greta for $300 making a 25% profit. D

Mashutka [201]

No Alan did not break even

Alan incured a loss of 6.25 %

Solution:

Given that,

Alan bought two bikes

He sold one to Beth for $300 taking a 25% loss

He also sold one to Greta for $300 making a 25% profit

When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss

Which is given as:

$loss\ \% = (\frac{x}{10})^2$

Here, x = 25

$loss\ \% = (\frac{25}{10})^2\\\\loss\ \% = 2.5^2\\\\loss\ \% = 6.25$

Thus the loss percentage is 6.25 %

8 0

3 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$