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Lelechka [254]
4 years ago
8

The coordinates are 5 -2 3-1 -4 4 -3 8 -14 find the shape with out plotting

Mathematics
1 answer:
lubasha [3.4K]4 years ago
8 0
Why don’t you draw that on a graph paper
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At which points does the graph of f(x)=2x−4 cross the x-axis and y-axis?
blagie [28]

crosses x-axis at (2, 0 ) and y-axis at (0, - 4 )

To find where the graph crosses the x and y axes ( intercepts )

• let x = 0, in the equation for y- intercept

• let y = 0, in the equation for x- intercept

x = 0 : y = 0 - 4 = - 4 ⇒ (0, - 4 )

y = 0 : 2x - 4 = 0 ⇒ 2x = 4 ⇒ x = 2 ⇒ (2, 0 )


6 0
3 years ago
5 cm<br> 4 cm<br> 6 cm<br> 3D rectangle
Sergio [31]

Answer:

120

Step-by-step explanation:

I'm assuming your looking for the area (if not pls tell me and I will make corrections)

So to get the area of a 3D shape you multiply the length, width, and height

5×4×6=120

8 0
3 years ago
if you repeat the perpendicular line segment construction twice using paper folding, you can construct:
ehidna [41]

Answer:

The correct answer is  option C.

The mid point of the line segment.

Step-by-step explanation:

the perpendicular line segment construction twice using paper folding

we have to find the mid point of the given line segment.

We get the midpoint easily when fold the paper correctly

Therefore the correct answer is  option C.

The mid point of the line segment.

6 0
3 years ago
Read 2 more answers
A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
Julli [10]

\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

5 0
2 years ago
can anyone help me asap plzz. Solve the inequality k + 6 ≤ 21; (A)k ≥ 27 (B)k ≤ 15 (C)k ≥ 15 (D)k ≤ 27
Ksenya-84 [330]

Answer:

Solve the inequality k + 6 ≤ 21;

(A)k ≥ 27

<u>(B)k ≤ 15 </u>

(C)k ≥ 15

(D)k ≤ 27

Step-by-step explanation:

K+6\leq 21\\    -6        -6\\K\leq 15

6 0
3 years ago
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