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3 years ago

17. Higher Order Thinking Solve h - 156 = 34

Mathematics

1 answer:

vaieri [72.5K]3 years ago

8 0

Answer: h = 190.

Step-by-step explanation: You have to use the reverse so, the opposite of subtraction is addition, so 156 + 34 = 190. 190 which is the "h" 190 - 156 = 34

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Bonjour je ne comprend pas cette exercice; si quelqu'un peut m'aider je suis preneuse.

gayaneshka [121]

Answer:

0,24 m

Step-by-step explanation:

La table est un rectanble donc l'angle en haut à droite est de 90°.

Tu connais donc un angle et le côté opposé à l'angle et tu cherche le côté adjacent.

On sait que tan(â) = $\frac{Cote oppose}{Cote adjacent}$ donc :

tan(40)= $\frac{1,27}{x}$

$\frac{1,27}{tan(40)} = x$

donc x ≈ 1,51 m

Le trou du mileu est à $\frac{2,54}{2}$ soit 1,27m du bord.

Donc elle doit taper à 1,51 - 1,27 = 0,24m du trou du milieu.

j'espère avoir été clair.

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2 years ago

9 - 3 ( 5 - 4 x ) =============

tangare [24]

Answer:26x will be your answer

8 0

3 years ago

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Rewrite as a product of two factors: 3 + 18d + 21y

Vlada [557]

Answer:

the Expression is :

(3) × (1 + 6d + 7y)

Step-by-step explanation:

here's the solution : -

=》

$3 + 18d + 21y$

=》

$(3 \times 1 )+ (3 \times 6d) + (3 \times 7y)$

now, since 3 is common in all terms let's take it out.

=》

$3 \times (1 + 6d + 7y)$

=》so, ,we got the above expression as product of 3 and ( 1 + 6d + 7y ) as factors.

3 0

3 years ago

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Mr.Church had 3 muffins he sold 2 but then baked 6 more and sold 4 more how many muffin did he not sell.

9966 [12]

He failed to sell three muffins

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3 years ago

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1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle

zimovet [89]

Answer: $\left | 2t-5\right |,\ 2,\ 2t-5$

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

$s(t)=t^2-5t+1$

Speed of the particle is given by

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left | 2t-5\right |$

Acceleration of the particle is

$\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2$

velocity can be negative, but speed cannot

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5$

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3 years ago