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3 years ago

Help me pls . this is alegbra 1

Mathematics

2 answers:

AURORKA [14]3 years ago

8 0

Answer:

x> -14/-3

Step-by-step explanation:

You start with the question -3x+4<-10

this you subtract 4 from both sides, now you have -3x<-14

now you need to isolait x by dividing both by -3x now you have x > -14/-3

I really hope this helps

Vanyuwa [196]3 years ago

6 0

Answer:

Step-by-step explanation:

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( please help, 20 points!! )

deff fn [24]

Measure of anfle 5=150

7 0

3 years ago

Triangle XYZ was dilated by a scale factor of 2 to create triangle ACB and cos ∠X = 2 and 5 tenths over 5 and 59 hundredths.

Maru [420]

Based on the above, the Triangle XYZ and ACB are known to be similar triangles since tanX = tanA = 5 over 2 and 5 tenths;

Hence:

AC = 2 x XY
CB = 2 x YZ

<h3>What is the triangle about?</h3>

From the question given; Triangle XYZ is known to have been dilated by a scale factor of 2 to form a type of triangle known as triangle ACB, So therefore, XYZ and ACB are known to be the same or similar triangles.

Since Angles Y and C are known to both measure 90 degrees, and angles A and X are said to be congruent. Therefore:

tanX = tanA = 5 over 2 and 5 tenths

AC = 2 x XY

CB = 2 x YZ

For part B:

cosX=Base/Hypotenuse

cos A=AC/AB

AC = 2.05

AB = 5.59

Therefore, Based on the above, the Triangle XYZ and ACB are known to be similar triangles since tanX = tanA = 5 over 2 and 5 tenths;

Hence:

AC = 2 x XY
CB = 2 x YZ

Learn more about triangles from

brainly.com/question/27517432

#SPJ1

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2 years ago

Please help! thank you!

anyanavicka [17]

Answer:

B. r=7.5

yan po ang correct answer

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2 years ago

The number of calories Kira burns while riding a bike varies directly with the number of hours the bike is ridden. How many calo

Veseljchak [2.6K]

480 since 1 hour=60 minutes. Look on the graph and you can see that the line looks like its 500 but since the answer choices dont say 500 it has to be 480 which is the closest

4 0

3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago