All categories

3 years ago

Please help please please

Mathematics

1 answer:

Natalka [10]3 years ago

6 0

He needs 5.1 more wired fence to get to 13.

You might be interested in

Factor 6p^2 +15p+9????

Ivanshal [37]

To factor the trinomial, you can use the ac method where you multiply the first and last terms together (6 and 9) and add up to 15.
In this case ac= 54
2 numbers that multiply to 54 and add up to 15 would be 6 and 9 so place these numbers where the 15 would be:
$6p^{2} +9p + 6p+9$
Then group terms to simplify:
$(6p^{2} +9p)(6p+9)
$
Simplify any like terms:
$3p(2p+3) 3(2p+3)$
And group together again:
$(3p+3)(2p+3)$
You can simplify a 3 out of the first expression one more time to get:
$3(p+1)(2p+3)$
Which would be the factored form. I hope this helps :)

8 0

3 years ago

Read 2 more answers

Help plz will give brainliest

astra-53 [7]

It should be D, I did the math :)

5 0

3 years ago

Read 2 more answers

Wxyz is a rectangle, what is zx

Alika [10]

ZX is a diagonal line from corner to corner.

6 0

3 years ago

Read 2 more answers

Help me please will be appreciated thx

alexandr1967 [171]

Answer:

-6 +8d -2c

Step-by-step explanation:

$\frac{-2}{5} (15 -20d +5c) = \frac{-2}{5} \times 15 - \frac{-2}{5}\times20d +\frac{-2}{5}\times5c = -6 + 8d -2c$

3 0

3 years ago

[4]

Tanya [424]

Answer:

Angle ACD = 38°

Step-by-step explanation:

The full, correct question is presented the attached image to this solution.

Given

Point O is the centre if the circle

Points A, B, C and D are points on the circle

Angle AOB = 140°

Angle OAC = 14°

Angle AOB = 2 × (Angle ACB) [angle subtended at the centre of the circle is twice the angle subtended at the circumference of the circle)

140° = 2 × (Angle ACB)

Angle ACB = (140°/2) = 70°

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180° [sun of angles in a triangle is 180°]

But Angle OAB = Angle ABO = a [base angles of an iscosceles triangle are equal since OA and OB are both radii for the circle O)

(Angle AOB) + (Angle OAB) + (Angle ABO) = 180°

140° + a + a = 180°

2a = 40°

a = (40/2) = 20°

Angle OAB = Angle ABO = 20°

Angle CAB = (Angle OAC) + (Angle OAB) = 14° + 20° = 34°

Triangle ADC is an iscosceles triangle, hence,

Angle DAC = Angle ACD = x [base angles of an iscosceles triangle are equal]

But

(Angle DCB) + (Angle DAB) = 180° [Opposite angles of a cyclic quadilateral (a quadilateral inscribed in a circle) sum up to give 180°]

Angle DCB = (Angle DCA) + (Angle ACB) = (x + 70°)

Angle DAB = (Angle DAC) + (Angle CAB) = (x + 34°)

(Angle DCB) + (Angle DAB) = 180°

(x + 70°) + (x + 34°) = 180°

2x + 104° = 180°

2x = 180° - 104° = 76°

x = (76°/2) = 38°

Angle DAC = Angle ACD = 38°

Angle ACD = 38°

Hope this Helps!!!

7 0

3 years ago