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alekssr [168]
2 years ago
12

What is the solution? (Look at picture)

Mathematics
1 answer:
max2010maxim [7]2 years ago
7 0

Answer:

(1,-1)

Step-by-step explanation:

The solution is where the two graphs cross.

The two lines cross at x=1 and y = -1

(1,-1)

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the height of an equivalent triangle is 15cm and it's perimeter is 36cm find the area of the triangle​
OlgaM077 [116]

Answer:

The area is 72 cm².

Step-by-step explanation:

Since it is equivalent triangle

h = b

so,

b + b + b (since all sides are equal)= 36 cm

3b = 36 cm

or, b = 36/3

so, b = 12 cm

so

area of triangle = (1/2)×b×h

= (1/2)×12cm×12cm

= 6cm × 12cm

= 72 cm²

4 0
2 years ago
Please help! I need the answer now!!!!!!!!!
rewona [7]

Answer:

Find the area by multiplying the base by the height. From the problem, the length (base) of the rectangle is 5 units. The height of the rectangle is

3\frac{1}{2}

inches. Multiply to find the area.

5 \times  3 \frac{1}{2}  = 17 \frac{1}{2}

The area of the rectangle is 17 1/2 units squared. The answer you provided is correct!

5 0
3 years ago
If you run a marathon of 26 1/5 miles in a time of 5 4/5 how far do you run in 1 hour<br><br>pls
alisha [4.7K]

First let's make both the miles and the hours ran improper fractions:

26+\frac{1}{5}=\frac{131}{5}\\ 5+\frac{4}{5}=\frac{29}{5}

So, to find our miles per hour, we have to divide the miles by the hours to get: \frac{\frac{131}{5}\text{miles}}{\frac{29}{5}\text{hours}}=\frac{131\text{miles}}{29\text{hours}} \\ \text{ Now we have to divide the top and the bottom by 29 to get our miles per hour so: }\\ \frac{\frac{131}{29}}{\text{hour}} \approx \frac{4.52 \text{miles}}{\text{hour}}

8 0
2 years ago
Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
Stella [2.4K]

Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

5 0
3 years ago
WILL GIVE A 5 STAR RATING A THANKS A BRAINLIEST and 12 points.
Sauron [17]

since there are 180 degrees in a triangle so :


65+72+y=180  add like terms

137 + y=180 now subtract 137

y= 180-137

y=43 degrees

4 0
2 years ago
Read 2 more answers
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