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3 years ago

Which postulate can be used to prove that the triangles are congruent?

Mathematics

2 answers:

Varvara68 [4.7K]3 years ago

5 0

Haven't been taught this; but from what I searched up; I think the postulate for proving these shapes' congruence is Hypotenuse/Leg. I have no exact idea however. Correct me if I'm wrong :/

leva [86]3 years ago

3 0

t is the SAS postulate the side angle side because they both have the 90 degree angle and the sides are the same coming of of the 90

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Based on data collected from its production processes, Crosstiles Inc. determines that the breaking strength of its most popular

alexandr1967 [171]

Answer:

16% of its popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch.

Step-by-step explanation:

We are given that the breaking strength of its most popular porcelain tile is normally distributed with a mean of 400 pounds per square inch and a the standard deviation of 12.5 pounds per square inch.

Let X = the breaking strength of its most popular porcelain tile

SO, X ~ Normal( $\mu=400,\sigma^{2}=12.5^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean breaking strength of porcelain tile = 400 pounds per square inch

$\sigma$ = standard deviation = 12.5 pounds per square inch

Now, probability that the popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch is given by = P(X > 412.5)

P(X > 412.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{412.5-400}{12.5}$ ) = P(Z > 1) = 1 - P(Z $\leq$ 1)

= 1 - 0.84 = 0.16

Therefore, 16% of its popular porcelain tile will have breaking strengths greater than 412.5 pounds per square inch.

7 0

3 years ago

Write the slope-intercept form of the equation for the line that passes through the point (-1,3) and has a slope of 5/3

cricket20 [7]

Answer:

$y=\frac{5}{3}x+4\frac{2}{3}$

Step-by-step explanation:

slope-intercept form: y = mx + b

Given:

Slope(m) = $\frac{5}{3}$

Point = (-1, 3)

To write the equation in slope-intercept form we need to know the slope(m) and the y-intercept(b). Since we already know the value of m, we can use it and the given point to find b:

$3=\frac{5}{3}(-1)+b$

$3=-\frac{5}{3}+b$

$4\frac{2}{3}=b$

Now that we know the values of b and m, we can write the equation:

$y = \frac{5}{3}x+4\frac{2}{3}$

4 0

2 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).