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2 years ago

Need the answer asap i will give you brainliest

Mathematics

1 answer:

ladessa [460]2 years ago

7 0

Answer:

Step-by-step explanation:

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I need help with a math problem...

Furkat [3]

1. x=5/2
2. x=3/4-3/4y
y belongs R

3 0

2 years ago

Robert went to the grocery store and purchased cans of soup and frozen dinners. each can of soup has 300 mg of sodium and each f

Paraphin [41]

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Step-by-step explanation:

Let,

Soup = x

Frozen Dinner = y

According to given statement of sodium;

300x + 500y = 5200 eqn 1

And,

x + y = 14 eqn 2

$Multiplying\ equation\ 2\ with\ 300\\300(x+y=14)\\300x+300y=4200\ Eqn 3\\$

Subtracting Eqn 3 from Eqn 1;

$(300x+500y)-(300x+300y)=5200-4200\\300x+500y-300x-300y=1000\\200y=1000\\y=\frac{1000}{200} \\y=5$

Putting value of y in Eqn 1;

$300x+500(5)=5200\\300x+2500=5200\\\\Subtracting\ 2500\ from\ both\ sides\ \\300x+2500-2500=5200-2500\\300x=2700\\x=\frac{2700}{300} \\x=9$

There are 9 cans of soup and 5 cans of frozen dinner.

The variables used for soup and frozen dinner are x and y respectively.

Keywords: Variables, linear equations, subtraction.

Learn more about linear equations at;

#LearnwithBrainly

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2 years ago

for a walk-a-thon a sponsor committed to give to give you a flat fee of $5 plus $2 for every Mile you walk write an expression f

oksian1 [2.3K]

You will get 57.44’$

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2 years ago

The line L is a tangent to the curve with equation y= 4x^2 +1 . The line L cuts the y axis at (0,8) and has a positive gradient.

sergeinik [125]

A generic point on the graph of the curve has coordinates

$(x, 4x^2+1)$

The derivative gives us the slope of the tangent line at a given point:

$f(x) = 4x^2+1 \implies f'(x) = 8x$

Let k be a generic x-coordinate. The tangent line to the curve at this point will pass through $(k, 4k^2+1)$ and have slope $8k$

So, we can write its equation using the point-slope formula: a line with slope m passing through $(x_0, y_0)$ has equation

$y-y_0 = m(x-x_0)$

In this case, $(x_0, y_0)=(k, 4k^2+1)$ and $m=8k$ , so the equation becomes

$y-4k^2-1 = 8k(x-k)$

We can rewrite the equation as follows:

$y-4k^2-1 = 8k(x-k) \iff y = 8kx - 8k^2+4k^2+1 \iff y = 8kx-4k^2+1$

We know that this function must give 0 when evaluated at x=0:

$f(x) = 8kx-4k^2+1 \implies f(0) = -4k^2+1 = 8 \iff -4k^2 = 7 \iff k^2 = -\dfrac{7}{4}$

This equation has no real solution, so the problem looks impossible.

5 0

2 years ago

What us probabilith<br>

julia-pushkina [17]

♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫

➷Probability is the chance of getting something randomly.

✽

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

TROLLER

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2 years ago