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3 years ago

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Mathematics

1 answer:

Feliz [49]3 years ago

6 0

Answer:

That's what I think the answer is

CB = CA - AB

= -8 - 29

= -37

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Need help with this math

ololo11 [35]

Answer:

Step-by-step explanation:

8 0

4 years ago

What are the factors of x2 – x – 6? Check all that apply.

Zina [86]

The factors would be (x-3)(x+2)

5 0

3 years ago

Write an exponential function y=ab^x whose graph passes through (1,6) and (3,54)

ehidna [41]

$y=ab^x\\\\(1;\ 6)\to x=1;\ y=6\\(3;\ 54)\to x=3;\ y=54\\\\\text{substitute the values of x and y to the equation}\\\\ \left\{\begin{array}{ccc}6=ab^1&\to&a=\dfrac{6}{b}^{(*)}\\54=ab^3\end{array}\right\\\\\text{substitute (*) to the second equation}\\\\54=\dfrac{6}{b}\cdot b^3\\\\54=6\cdot b^2\ \ \ \ |:6\\\\b=9\to b=\sqrt9\to b=3 \\\\\text{substitute the value of b to the (*)}\\\\a=\dfrac{6}{3}=2\\\\\text{Answer:}\ y=2\cdot3^x$

7 0

3 years ago

Eliminate the parameter for the following set of parametric equations: x= t^2 + 2 y= 4t^2

mr_godi [17]

Answer:

Solution : y = 4x - 8

Step-by-step explanation:

The first thing we want to do is isolate t², rather than t. Why? As you can see when we substitute t² into the second equation, it will be easier than substituting t, as t is present in the form t². So, let's isolate t² in the first equation --- ( 1 )

x = t² + 2,

t² = x - 2

Now let's substitute this value of t² in the second equation --- ( 2 )

y = 4t²,

y = 4(x - 2),

y = 4x - 8 ~ And hence our solution is option c.

6 0

3 years ago

Solve the given system of equations utilizing either the substitution or addition method.

Nataly [62]

Answer:

Value of x=4 and y=-13/3

Step-by-step explanation:

We need to solve the systems of equations

$5x+3y=7 \ and \ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}$

Solving using substitution method:

Let:

$5x+3y=7---eq(1) \\ \frac{3}{2}x-\frac{3}{4}y =9\frac{1}{4}---eq(2)$

Find value of x from eq(1) and put in eq(2)

$5x+3y=7\\5x=7-3y\\x=\frac{7-3y}{5}$

Put value of x in eq(2)

$\frac{3}{2}(\frac{7-3y}{5}) -\frac{3}{4}y =\frac{37}{4}\\\frac{21-9y}{10} -\frac{3}{4}y =\frac{37}{4}\\Taking \ LCM\\\frac{21*2-9y*2-3y*5}{20} =\frac{37}{4}\\\frac{42-18y-15y}{20} =\frac{37}{4}\\42-33y=\frac{37}{4}*20\\42-33y=185\\-33y=185-42\\-33y=143\\y=\frac{143}{-33}\\y=-\frac{13}{3}$

Now, finding value of x by putting value of y in eq(1)

$x=\frac{7-3y}{5}\\x= \frac{7-3(-\frac{13}{3}) }{5}\\x=\frac{7+13}{5}\\x=\frac{20}{5}\\x=4$

So, Value of x=4 and y=-13/3

4 0

3 years ago