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3 years ago

10

Which equation shows y=34x−52 in standard form?

1 answer:

Dafna1 [17]3 years ago

6 0

The answer would be a 3x-4y=-10

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Please help me with this problem

oksian1 [2.3K]

Diagonal = SQRT(8^2 + 12^2 + 30^2)

Diagonal = SQRT(64 + 144 + 900)

Diagonal = SQRT(1108)

Diagonal = 33.3 feet

5 0

3 years ago

Quanto fa 2×2? hsbwhannwn

goldenfox [79]

Answer:

$ok, ma \: quando \: 2+2= 1? $

6 0

2 years ago

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Please help me <br><br>thank you

Sveta_85 [38]

Answer:

M

Hope this helps!

Take care!

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2 years ago

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Enter an equation for the function that includes the points.Give your answer in a(b)x. In the event that a=1 , give your answer

Andrews [41]

Answer:

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

Step-by-step explanation:

Given

$(x_1,y_1) = (2,\frac{2}{3})$

$(x_2,y_2) = (3,\frac{5}{9})$

Required

Write the equation of the function $f(x) = ab^x$

Express the function as:

$y = ab^x$

In: $(x_1,y_1) = (2,\frac{2}{3})$

$y = ab^x$

$\frac{2}{3} = a * b^2$ --- (1)

In $(x_2,y_2) = (3,\frac{5}{9})$

$y = ab^x$

$\frac{5}{9} = a * b^3$ --- (2)

Divide (2) by (1)

$\frac{5}{9}/\frac{2}{3} = \frac{a*b^3}{a*b^2}$

$\frac{5}{9}/\frac{2}{3} = b$

$\frac{5}{9}*\frac{3}{2} = b$

$\frac{5}{3}*\frac{1}{2} = b$

$\frac{5}{6} = b$

$b = \frac{5}{6}$

Substitute 5/6 for b in (1)

$\frac{2}{3} = a * b^2$

$\frac{2}{3} = a * \frac{5}{6}^2$

$\frac{2}{3} = a * \frac{25}{36}$

$a = \frac{2}{3} * \frac{36}{25}$

$a = \frac{2}{1} * \frac{12}{25}$

$a = \frac{24}{25}$

The function: $f(x) = ab^x$

$f(x) = \frac{24}{25} * \frac{5}{6}^x$

7 0

3 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$

And so the equation is

$x^2 - \dfrac{5}{4} + \dfrac{3}{8} = 0$

In order to have integer coefficients, you can multiply both sides of the equation by 8:

$8x^2 - 10 + 3 = 0$

5 0

3 years ago