3 years ago

I need help asap (show the work)

Mathematics

1 answer:

MArishka [77]3 years ago

3 0

Answer:

X = 12

Is the answer,

Happy

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Ilia_Sergeevich [38]

Answer:

B= 40 degress

Step-by-step explanation:

1x+ 5xc /3xc = 40

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2 years ago

A^2-b^2 a^2b-ab^2<br>-------------- ÷ ---------------a^2b+ab^2 a^2b- ab^2

Damm [24]

Answer:

$\huge{= \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} }}$

Step-by-step explanation:

$\huge{ \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} } \div \frac{ {a}^{2b} - {ab}^{2} }{ {a}^{2b} - {ab}^{2} }}$

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$\huge{\boxed{\green{= \frac{ {a}^{2} - {b}^{2} }{ {a}^{2b} + {ab}^{2} }}}}$

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3 years ago

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Bezzdna [24]

Answer:

$3.41

Step-by-step explanation:

262.94 + 83.65 = 346.59

350.00 - 346.59 = 3.41

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3 years ago

Can someone do this?

malfutka [58]

Answer:

Step-by-step explanation:

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2 years ago

The area of triangle PQR is 273 cm. Given that PQ = 12.8cm and PQR = 107°, find QR.

storchak [24]

$\text{Area of triangle} = \dfrac 12 \times PQ \times QR \times \sin \angle PQR \\\\\\\implies 273= \dfrac 12 \times 12.8 \times QR \times \sin 107^{\circ}\\\\\implies 273 = 6.4 \sin 107^{\circ} \times QR\\\\\\\implies QR = \dfrac{273}{6.4 \sin 107^{\circ}} = 44. 606~~ \text{cm}$

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2 years ago