All categories

2 years ago

The triangles shown are congruent.

Mathematics

2 answers:

lbvjy [14]2 years ago

5 0

Answer: SSS congruence postulate

Step-by-step explanation:

In the given picture , we have two triangles $\triangle{QDJ}\text{ and}\triangle{MCW}$ .

In $\triangle{QDJ}\text{ and}\triangle{MCW}$ , we have

$\overline{QJ}\cong\overline{MW}\\\\\overline{DJ}\cong\overline{CW}\\\\\overline{QD}\cong\overline{MC}$

Then by SSS congruence postulate , we have

$\triangle{QDJ}\cong\triangle{MCW}$

SSS congruence postulate says that any two triangles are congruent if all three sides in first triangle are congruent to the corresponding sides in the second.

cestrela7 [59]2 years ago

4 0

The correct congruence statement is SSS, or Side-Side-Side. We see that lines QD and MC are equivalent, QJ and MW are equivalent, and lines DJ and CW are equivalent

You might be interested in

What are the new vertices of triangle QRS, shown, if the triangle is reflected across the line y = x, also shown? Question 6 opt

ExtremeBDS [4]

Answer:

Step-by-step explanation:

Under a reflection in the line y = x

a point (x, y ) → (y, x ) , thus

Q(- 5, 2 ) → Q'(2, - 5 )

R(0, 5 ) → R'(5, 0 )

S(- 1, 2 ) → S'(2, - 1 )

6 0

3 years ago

Which following figures have a area of 64 square feet? Select three that apply

Korvikt [17]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find the key features of the equation and graph it. (Can you explain how to find them and not just the answers please)

Vadim26 [7]

Given a function in a table or in algebraic or graphical form, identify key features such as x- and y-intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. Use key features of an algebraic function to graph the function.

7 0

2 years ago

I need help with all three letters. thank u

creativ13 [48]

Answer:

Step-by-step explanation:

A to C

y = b + mx

b = y-intercept

m = slope of line

Looking at the line, the line crosses the y-axis at the coordinates (0,0). This means that the y-intercept is 0.

The slope formula is m=(y2-y1)/(x2-x1). Using the points on the line (-1, -3) and (1,3) we find that the slope is, 3.

Looking back we can see that our variables now have value so we can plug them into our formula.

y = b + mx

b = 0

m = 3

Substitute

y = 0 + 3x

4 0

2 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

2 years ago

Read 2 more answers