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3 years ago

A.SAS Postulate

Mathematics

1 answer:

tankabanditka [31]3 years ago

8 0

Answer:

C. you can not prove that the triangles are congruent

Step-by-step explanation:

The answer can not be SAS becuase although the congruent angle is not in between the two congruent sides.
It can not be AAS because only one angle is congruent.
It can not be SSS because all three sides are not congruent.

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Solve for x: 2(x 3)2 − 4 = 0 Round your answer to the nearest hundredth. X = 4. 41, 1. 59 x = 1. 34, 5. 24 x = −1. 34, −5. 24 x

gladu [14]

The two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

<h3>What is a quadratic equation?</h3>

A quadratic equation is the equation in which the unknown variable is one and the highest power of the unknown variable is two.

The standard form of the quadratic equation is,

$ax^2+bx+c=0$

Here, (<em>a,b,c</em>) are the real numbers and <em>x </em>is the variable.

To find the value of x, the following formula is used,

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The given equation is,

$2(x+ 3)^2 - 4 = 0$

To solve this equation, we need to apply some mathematical operations over it. Let's start with opening the brackets.

$2(x+ 3)^2 - 4 = 0\\2(x^2+6x+9)-4=0\\2x^2+12x+14-4\\2x^2+12x+12=0\\x^2+6x+7=0$

On comparing with standard equation we get,

$a=1, b=6, c=7$

Put this values in the above formula,

$x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=\dfrac{-(6)\pm\sqrt{(6)^2-4(1)(7)}}{2(1)}\\x=-4.41,-1.59$

Hence, the two values when the provided quadratic equation is solved for the x are -4.41 and -1.59 to the nearest hundredth.

Learn more about the quadratic equation here;

brainly.com/question/1214333

5 0

2 years ago

If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)

Eva8 [605]

We can expand the logarithm of a product as a sum of logarithms:

$\log_dabc=\log_da+\log_db+\log_dc$

Then using the change of base formula, we can derive the relationship

$\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}$

This immediately tells us that

$\log_dc=\dfrac1{\log_cd}=\dfrac12$

Notice that none of $a,b,c,d$ can be equal to 1. This is because

$\log_1x=y\implies1^{\log_1x}=1^y\implies x=1$

for any choice of $y$ . This means we can safely do the following without worrying about division by 0.

$\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}$

so that

$\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23$

Similarly,

$\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}$

so that

$\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34$

So we end up with

$\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}$

###

Another way to do this:

$\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}$

$\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}$

$\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12$

Then

$abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}$

So we have

$\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}$

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2 years ago

What is the probability that two people in a class of 32 will have the same birthday?

mamaluj [8]

2/32 - 1/16 this is the probability (in fractions)

4 0

3 years ago

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The digits 1-5 are used for a set of locker codes. Suppose the digits cannot repeat. Find the number of possible two digit and t

Archy [21]

So you would do
5×4×3= 60
and
5×4=20
so, for a 3 digit code there are 60 possibilities and for a 2 digit code there are 20 possibilities.

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3 years ago

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Write the equation of the line that passes through (-2, 6) and (2, 14) in slope-<br> intercept form.

creativ13 [48]

Answer:

Y= 2x+10

Step-by-step explanation:

(-2, 6), (-1, 8), (0, 10), (1, 12), (2, 14)

You add 2 every time

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2 years ago