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3 years ago

I WILL GIVE BRAINLIEST FAST

Mathematics

1 answer:

Sloan [31]3 years ago

8 0

Answer:

is opposite line BC. Answer is letter E.

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Please help!!!!!!!!!!!!!!!!!!!!

grigory [225]

Answer:

f(3)=g(6) point of intersection

3 0

3 years ago

Which shows the correct way to use the rule "multiply by the reciprocal" for the expression below? 5 divided by StartFraction 2

enot [183]

Answer:

its a.

Step-by-step explanation:

simple answer

5 0

3 years ago

Read 2 more answers

Please help me ASAP!!!!

Ket [755]

ANSWER

$34 {units}^{2}$

EXPLANATION

The width of the rectangle is |AB|

A(-1,4) and B(3,3)

We use the distance formula,

$d = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

This implies that,.

$d = \sqrt{ {(3- - 1)}^{2} + {(3 - 4)}^{2} }$

$d = \sqrt{ {(4)}^{2} + {( - 1)}^{2} }$

$d = \sqrt{ 16+ 1 }$

$d = \sqrt{17}$

The length is BC

C(1,-5) and B(3,3)

$d = \sqrt{ {(3- 1)}^{2} + {(3 - - 5)}^{2} }$

$d = \sqrt{ {2}^{2} + {(8)}^{2} }$

$d = \sqrt{4 + 64 }$

$d = \sqrt{68}$

$d = 2 \sqrt{17}$

The area is

$= l \times w$

$= 2 \sqrt{17} \times \sqrt{17}$

$2 \times 17 = 34 {units}^{2}$

5 0

4 years ago

Fill out the following chart to find the temperatures for t = 12 (noon) and t = 24 (midnight). (2 points: 1 point for each row)

sergey [27]

Part.1
Given equation:
C(t) = -0.30 (t – 12)² + 40

For t = 0
C(t) = -0.30 (0 - 12)² + 40
C(t) = -0.30 (-12)² + 40
C(t) = -3.2

For t = 12 (noon)
C(t) = -0.30 (12 - 12)² + 40
C(t) = -0.30 (0)² + 40
C(t) = 40

For t = 24 (midnight)
C(t) = -0.30 (24 - 12)² + 40
C(t) = -0.30 (12)² + 40
C(t) = -0.30 × 144 + 40
C(t) = - 43.2 + 40
C(t) = -3.2

Part.2
Graph of the equation is given in the attachment.

Part.3
C(t) = –0.30(t – 12)² + 40
F(t)=9/5C(t)+32

Substituting the values:
F(t)=9/5{–0.30 (t – 12)² + 40}+32
F(t) = -0.54 (t – 12)² + 72 + 32
F(t) = -0.54 (t – 12)² + 104

6 0

3 years ago

Read 2 more answers

A student solved the following problem and made an error: Triangles ABC and DEF. Angles A and F are congruent and measure 135 de

nignag [31]

The first false statement in the proof as it stands is in Line 5, where it is claimed that a line of length 2.83 is congruent to a line of length 4.47. This mistake cannot be corrected by adding lines to the proof.

_____

The first erroneous tactical move is in Line 4, where the length of DE is computed, rather than the length of FD. This mistake can be corrected by adding lines to the proof.

A correct SAS proof would use segment FD in Line 4, so it could be argued that the first mistake is there.

7 0

3 years ago