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3 years ago

Plz help will mark the brainliest

Mathematics

1 answer:

Alisiya [41]3 years ago

4 0

Step-by-step explanation:

Rana Pratap Sagar Dam

b. Namrup Thermal Plant

c. Bengaluru Software Technology Park

d. Vishakhapatnam Port

e. Naraura Nuclear Power Plant

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Sheldon also picks tomatoes from his garden. He picked 4 4/10 but 1.5 were rotten and had to be thrown away. How many kilograms

velikii [3]

Given:

Number of tomatoes Sheldon picks = $4\dfrac{4}{10}$ kg

Rotten tomatoes = 1.5 kg

To find:

How many kilograms of tomatoes were not rotten?

Solution:

Number of tomatoes Sheldon picks = $4\dfrac{4}{10}$ kg

= $\dfrac{4(10)+4}{10}$ kg

= $\dfrac{40+4}{10}$ kg

= $\dfrac{44}{10}$ kg

= $4.4$ kg

Rotten tomatoes = 1.5 kg

Remaining tomatoes which are not rotten is

Required tomatoes = Tomatoes Sheldon picks - Rotten tomatoes

= $4.4-1.5$

= $2.9$ kg

Therefore, the tomatoes which were not rotten is 2.9 kg.

8 0

3 years ago

Would the answer be c??

andrew-mc [135]

Answer:

<h2>PLEASE MARK BRAINLIEST!!</h2>

Yes

3 0

3 years ago

Transformation for y=-2(x-1)2 + 2

Llana [10]

Answer:

Vertex = 1,2

it is shifted 2 units up and to the right 1 unit

it is a downward parabola (max at (1,2) going down

Step-by-step explanation:

5 0

3 years ago

Simplify the equation 3×5x

Setler79 [48]

Answer:

15x

3×5x

3×5 =15

15*x

15x

hope it's helpful ❤❤❤❤❤❤

THANK YOU.

7 0

3 years ago

Read 2 more answers

Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

3. $\mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1$

4. $\mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}$

5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

3 years ago

Read 2 more answers