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3 years ago

HELP PLEASE LOTS OF POINTS

Mathematics

1 answer:

inna [77]3 years ago

8 0

We are given

p: 1 is the multiplicative identity

Firstly, we will verify whether it is true

Multiplicative identity:

$A \cdot 1=A$

Here , 1 is the multiplicative identity

So, 1 is multiplicative identity

so, statement p is true

q: a · 0 = a

we know that whenever we multiply any expression by 0

It will result in 0

so, this statement is false

now, we will verify each options

option-A:

p \/ q

so, we can plug

T \/ F = T

so, this is TRUE

option-B:

p /\ q

so, we can plug

T /\ F = F

so, this is FALSE

option-C:

p --> q

we can plug

T-->F = F

so, this is FALSE

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vlabodo [156]

Answer : a
Sin is opposite/hypotenuse which is 8/10
And this is 4/5

3 0

3 years ago

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Kryger [21]

Answer:

Given the triangle RST with Coordinates R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.

Since, the scale factor $\frac{5}{3}$ is greater than 1, the image is enlargement or a stretch.

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor $\frac{5}{3}$

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image $S{}'$ are same.

Now, the distances from the center of the dilation at point S to the other points R and T.

The dilation image will be $\frac{5}{3}$ of each of these distances,

$SR=3$ , so $S{}'R{}'$ =5 ;

$ST=3$ , so $S{}'T{}'=5$

Now, draw the image of RST i.e R'S'T'

Since, $RT=3\sqrt{2}$ [By using hypotenuse of right angle triangle] and $R{}'T{}'=5\sqrt{2}$ .

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of $\overline{AC}$ i.e $\overline{AB}\cong \overline{BC}$

In the triangle ABD and triangle CBD, we have

$\overline{AB}\cong \overline{BC}$ (SIDE) [Given]

$\overline{BD}\cong \overline{BD}$ (SIDE) [Reflexive post]

Since, there is no included angle in these triangles.

∴ $\Delta ABD$ is not congruent to $\Delta CBD$ .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that $\overline{AD}\cong \overline{CD}$ .