Which equation has no soultion

Slope-intercept form (-3,-4);=-5x+5

Harlamova29_29 [7]

Answer:

with the way you wrote it, it's unidentified

8 0

2 years ago

A blank is a rectangular

UNO [17]

Answer:

I would say parallelogram. So the rectangle still fits but more shapes can be applied.

8 0

3 years ago

Given a scalene triangle with whole number angle measurements, what is the greatest sum possible of the two smallest angles?

Olegator [25]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Define scalene triangle

A Scalene Triangle is any triangle with unequal sides. This means that, in a scalene triangle, all of the three sides and angles are different lengths, just like in the illustration below. This also means that each angle has to be different.

STEP 2: Get the greatest sum possible of the two smallest angles

Since the measures of the angles of the triangle are different whole numbers , for the sum of the two angles to be the least possible, one of the angles must be smallest whole number i.e 1°

Now, the next smallest angle will be the next angle of a whole number that is not 1 degree, i.e, 2 degrees.

Thus, the greatest possible sum of the measures of two smallest angles will be:

$1^{\circ}+2^{\circ}=3^{\circ}$

Hence, the greatest sum possible of the two smallest angles is 3 degrees

7 0

1 year ago

HELP!!!!

m_a_m_a [10]

Answer:

(0,0)

Step-by-step explanation:

The coordinates of C now are (-7,-5).

Moving C -3 on the x-axis will move -7 - 3 = -10. C is now located on (-10,-5).

Moving C +5 on the y-axis will move -5 + 5 = 0. C is now located on (-10,0).

Moving C on the y-axis +10 will move -10 + 10 = 0. C is now located on (0,0).

C is translated to (0,0).

8 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

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