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2 years ago

7

Please help . In the figure, ∆BAT ≅ ∆CAT. Which statement is not true by CPCTC?

1 answer:

masya89 [10]2 years ago

3 0

Answer:

$\overline{CS}\cong\overline{ST}$

Step-by-step explanation:

The congruence statement for the triangles does not list point (or angle) S, so no claim about segments or angles involving S will be supported by the congruence statement.

The statement

$\Delta BAT\cong\Delta CAT$ tells you ...

$\angle TBA\cong\angle TCA\\\angle BAT\cong\angle CAT\\\angle ATB\cong\angle ATC\\\overline{BA}\cong\overline{CA}\\\overline{AT}\cong\overline{AT}\\\overline{TB}\cong\overline{TC}$

No other claims can be made using CPCTC. The statement among the answer choices that is not true by CPCTC is ...

$\boxed{\overline{CS}\cong\overline{ST}}$

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Can someone please help me answer this?

Lina20 [59]

DEF = BAC : x = 15 <==

once u allign ur triangles, u will see that there is a scale factor of 1.5.
6 * 1.5 = 9

1.5(x - 5) = 15
1.5x - 7.5 = 15
1.5x = 15 + 7.5
1.5x = 22.5
x = 22.5/1.5
x = 15 <==

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2 years ago

What is the vertex form of Y = x2 - 6x + 6?

ahrayia [7]

Answer:

$\large\boxed{y=(x-3)^2-3}$

Step-by-step explanation:

The vertex form of a quadratic equation <em>y = ax² + bx + c:</em>

<em>y = a(x - h)² + k</em>

(h, k) - coordinates of a vertex

We have the equation <em>y = x² - 6x + 6</em>.

Convert to the vertex form:

$y=x^2-2(x)(3)+6\\\\y=\underbrace{x^2-2(x)(3)+3^2}_{(*)}-3^2+6\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\y=(x-3)^2-9+6\\\\y=(x-3)^2-3\to h=3,\ k=-3$

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2 years ago

The box-and-whisker plot below shows the age distribution of people at a family reunion.

Wewaii [24]

Answer: answer is c

Step-by-step explanation: quizlet

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3 years ago

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

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3 years ago

If A c B and A N B= • then which of the following can be concluded about the sets A and B!

Sergio [31]

Step-by-step explanation:

Set A is a subset of set B if all the elements of A are also the elements of Set B.. (common element)

The above set have 0 common elements so.. if there is no common element and are subsets then it means that there arent any elements in both the sets

so, the answer for the question would be "both the Sets A and B are the empty sets"

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1 year ago